TLDR:
No, you cannot use those batteries to power that motor. Nothing bad will happen if you try, but the safety cutoffs will engage the first time you try to turn it on and it just won't work.
If your requirement demands more energy density and energy per kilogram than lead-acid can provide, your only option is to use high discharge rate LiPo cells.
By 'full load', I assume you are referring to some sort of information placard on the motor itself that provides this information, and perhaps it is a motor specific to some sort of more or less fixed mechanical load?
DC motors do not draw a fixed current, but rather draw current inversely proportional to their rotation speed. The motor's rotor coils spin relative to the stator magnets, which means from the coils perspective, there is a constantly changing magnetic field. This of course induces a voltage in the coil - one that is opposite the voltage being applied to it by an external power source. The motor will spin faster and faster until the induced EMF nearly cancels out the voltage from, say, a battery. The rotor coils in an unloaded motor might have 200V applied externally, but draw far less current than the coil impedance could account for alone at that voltage. This is because the rotor coils don't have 200V across them. They might have as little as 1V across them, or less, or more (totally depends on the motor, just illustrating my point), because it is spinning vast enough that 199V of EMF is induced in the coils, 199V always in the opposing polarity to the power source.
This also means the torque is very low as well, as it is entirely generated magnetic fields, which in turn are generated by the magnetomotive force (MMF), current. When you load the shaft however, several things happen. The shaft will not have the torque to counter the turning resistance to the shaft, so the shaft slows down. It will slow, the induced EMF will fall, and the true voltage across the rotor coils will increase, and so too will the current drawn by the coils. The additional current drawn results in stronger magnetic fields which increase the torque the motor can push back with. Indeed, it will slow until the current necessary to perfectly counter the shaft resistance is being produced. Or, if the load is simply too much for the motor to overpower, the shaft will stop entirely, and no counter EMF is induced in the coils at all. The full brunt of the driving voltage is felt, and in general, a massive amount of current will be drawn by the motor. This will continue until something gives. Either the shaft begins turning, the motor's windings melt, or power is removed (many larger motors have a thermal cut off switch which does this by cutting power if the motor gets to hot).
The takeaway here is that motors are neither simple nor well-behaved loads. The only mechanism controlling current draw is the counter EMF. When a motor first turns on, there is none, as it is not spinning. The motor will draw its stall current which is both typically many many times its no load current, and still a multiple of it's 'loaded' current.
So you need to find out what you really mean by 'full load'. Often, if the motor is meant for a specific purpose, or is continuous duty, 'full load' means the maximum continuous current consumption that is allowable for that motor so it doesn't over heat. Motors generally specify other currents, specifically stall current, which is the current drawn when the shaft is not moving ('stalled') and no load current, which is the current drawn when there is nothing but friction/air resistance/etc loading an uncoupled bare shaft. No motor is designed (or probably able) to be used stalled, so it's fairly certain 'full load' is not referring to the stall current, but the current drawn at some point of shaft loading that the motor can manage indefinitely.
So no, you can't power your motor with those batteries, regardless of configuration.
The motor will not draw 21A maximum. It will draw 21A at it the maximum load it can operate it continuously. It will several times this current when first started up because, initially, the shaft is not moving and the motor is at stall. This current will be brief, but not nearly brief enough. The LiIon's protection circuits will cut power, and it will only take one battery doing this to cause an overload cascade that makes them all cut power. And they should - those protection circuits are in place and operate as fast as they do because drawing such a load can cause localized spot heating of the battery anode which could potentially make a hotspot above 80 °C in the adjacent solid electrolyte interphase (SEI) layer. Once that happens, a LiIon cell generally goes into thermal runaway and the battery converts itself to fire. Lots and lots of fire.
Don't worry, the protection circuits are very good at preventing that, to the point the LiIon cells have the lowest incident rate of any chemistry (as the other chemistries do not have protection circuits generally).
But they will certainly be overloaded by that motor, upon start at the least, and any time the motor has some transient load that makes it draw a bit too much current.
If you must power such a beast from batteries and lead-acid isn't energy dense enough for your application, your options are pretty much limited to high discharge rated LiPo cells. Thanks to the RC/quadrotor popularity, which require very high drain rates (because, well, that's what motors do), there is an abundance of such batteries readily available and it very reasonable cost. Just make sure to use a reputable vendor, many of the RC batteries give ...optimistic... numbers for their max continuous discharge rate.
These batteries still have protection circuits, but ones paired to the cells, and will only cut off if the maximum discharge current is exceeded, just like your camera batteries. But, that maximum can be in the hundreds of amps, depending on cell size.
Regardless, your application is a high-drain motor application, and you must use batteries that are designed for such a load, which by definition camera batteries are not. Mostly because cameras are not 4200W motors.
Additionally, it is not acceptable to put unmatched cells in series like that. The cells have lead different lives, are at different stages of wear and capacity, and with so many cells in series, it the are all sorts of other mechanisms that will cause the protection circuits to engage when the weakest cell in the weakest pack has discharged too much, even if there is plenty of capacity left in the rest of the pack.
You need to buy new, matched cells when putting them in series. Be sure to use a proper protection circuit for each. In general, it is good to avoid paralleling cells as well, though it can be done if needed. In your case, it is not, there are a plethora of LiPo cells able to handle the current demands of that motor, even at stall, but they still must be matched. And you'll need about 48 of them in series, which is certainly not an impractical stack. You will also need to a cell balancer and fairly advanced charger to maintain these cells.
If you create the pack by combining many smaller packs in series, which is probably the cheapest option (using hobby lipos with high enough discharge ratings), you will still need to actively balance the entire pack at once, so no charging them separately. You need a 200V charger/balancer. These too are readily available, as EV chargers. This one for example.
None of this is cheap, but your options are spending less money by buying new cells of the right ratings and the correct charging/balancing equipment for a 48S stack, or you can spend a fair bit more money doing it the wrong way, and when that doesn't work, also spending the money to do it the right way as well. So don't misunderstand, the option where you buy a $250 balancer and who knows how many dollars of new, prime high C rate lipo cells is the cheap option. The option that seems cheaper at first is the much more expensive option.
Go with the cheap option.
Best Answer
You don't say what you use the batteries for, and if you are running your motor at full speed all the time. It's often better to have more voltage than you need and then modulate or regulate the current to run over a longer period of time. Then as the battery voltage decays you can go longer without recharging. A d.c. motor will rotate at a speed creating a counter voltage that equals the source voltage. Often motors are under some form of feedback, like a servo amplifier. I have controlled high inductance motors that put power back into the control amplifier. That extra energy needs to be put somewhere when slowing down the motor - it acts like a generator. Extra energy needs to be released as heat, or put back into the powering battery. You don't need any extra voltage as long as you can run at the speed you desire. Don't forget that voltage times current equals force, so can you usefully use extra torque? Do you mind accelerating faster than you have before? The insulation on a d.c. motor is at least twice the rated voltage, usually much more. The constraints have more to do with force, top speed and control. Lighting a lamp with overvoltage will burn it out sooner, but that is not usually the main constraint in motors.