The first which comes to mind is bilateral switches like 74HC4066. They have a few tens of Ohms of on-resistance, but that won't probably much of a problem if your feedback resistor values are high enough, otherwise take them in account, and for example use a 950\$\Omega\$ resistor instead of a 1k.
I'm getting some rather negative reactions on this, but I just wanted to make the same remark endolith makes: if Matt can use a symmetrical power supply the switch can be placed on the side of the virtual ground, and especially when using a high feedback resistance there will be hardly any voltage drop over it, so also no distortion. If the feedback resistance is 10k\$\Omega\$ then the resistance of the switch is only 1/10th of the tolerance on a 1/4W E12 resistor.
FakeName suggests some AD devices as better alternatives to the 74HC4066. Particularly the ADG1401 he mentions looks very interesting, with an \$R_{ON}\$ < 2\$\Omega\$. (Thanks, Fake)
Matt already seems to have found Maxim analog multiplexers, but I want to include the MAX4781, which has similar \$R_{ON}\$, in case others are interested. The Intersil ISL84781 is a pin-for-pin replacement for the MAX4781, and has a maximum \$R_{ON}\$ of 0.8\$\Omega\$.
High end audio equipment sometimes uses reed-relays here, which have the advantages of a mechanical switch: m\$\Omega\$ resistance when on, G\$\Omega\$ when off.
edit2
Just had another idea! :-) An OTA (transconductance amplifier) can be used to make a VCA (Voltage Controlled Amplifier). This one uses an LM13700:
Instead of having a series of digital control lines controlling switches you vary the amplification with an analog voltage. DAC can be as simple as R-2R ladder network.
edit3
Yet another solution also takes the problem one level up, like the OTA: you don't really want switches, you want a variable gain amplifier, and therefore the feedback resistor must be variable. Like with a digital potentiometer. This one has a 100k\$\Omega\$ range in 400\$\Omega\$ steps. You can also choose a smaller range in combination with a fixed resistor, of course. Possible drawback here is that the wiper has a rather high resistance of 800\$\Omega\$ (maximum), though this just means you'll always have the two lowest steps.
You are trying to use the "opamp out of the range it behaves well". Used open loop like this it becomes a makeshift comparator.
Opamps are meant to always be used closed loop. The open loop gain is pretty variable so cannot be relied upon, the only real requirement is that it's as large as possible (ideally infinite) When you close the loop though the huge gain makes the (closed loop) gain very accurate and the open loop variability (usually) makes almost no difference.
In theory, with an ideal opamp what you are trying would work okay, but in real opamps you have slight mismatches, causing things like input offset voltage which just by itself will cause this experiment to fail.
If we look at page 3 of the datasheet, we see the typical offset voltage for the LM324 is 2mV, up to a maximum of 7mV (3mV for the LM324A). This swamps your tiny input signal, and will cause the output to do just about anything, probably saturate.
There is also an offset current, external noise, and other non-ideal parameters that are variable with temperature that will cause issues here.
For an example of how variable the open loop gain is, here is a graph of the open loop gain over supply voltage:
We can see at 5V, the open loop gain is around 100dB (100,000) compared to 120dB (1,000,000) at 15V. This is a factor of 10 difference. Also we can see it varies considerably with load.
Best Answer
To give you a better understanding what is going on in the inverting amplifier let us at the beginning use this circuit:
We simply have an ideal voltage amplifier with different \$A_O\$ gains.
Now let as try to find an input resistance.
$$R_{IN} = \frac{V_{IN}}{I_{1}}$$
$$I_{1} = \frac{V_{IN} - V_{OUT}}{R_F} =\frac{V_{IN} - A_OV_{IN}}{R_F} = V_{IN} \frac{1 -A_O}{R_F} $$
$$R_{IN} = \frac{R_F}{1-A_O}$$
And for inverting input we have
$$R_{IN} = \frac{R_F}{1-( - A_O)} = \frac{R_F}{1+|A_O|}$$
For example if \$A_O = 10\$ (open loop gain) in input current is \$I_1 = \frac{11V}{10k\Omega} = 1.1mA \$ and the input resistance is :
$$ R_{IN} = \frac{1V}{1.1mA} = 909.09\Omega$$
As you can see our \$R_{IN}\$ resistance is \$(1 + |A_O|)\$ smaller than \$R_F\$ if we have inverting amplifier. And this is what we call a Miller effect.
And now if we add a resistor between the signal source and the op-amp inverting input we created the inverting amplifier.
As you can see this time the amplifier input resistance is equal to:
$$R_{IN} = R_1 + \frac{R_F}{1+|A_O|}$$
And for the Op-amp \$A_O \to \infty \$ approach to infinity.
Hence we have a "virtual Ground" at the inverting input and \$R_{IN} = R_1\$ and the voltage gain is \$A_V = -\frac{R_F}{R_1}\$ due to negative feedback.
Or we can treat this circuit as voltage to current converter \$R_1\$ and the ideal Op-amp plus \$R_F\$ forms \$R_1\$ current to voltage converter.
The positive feedback (RF resistor between the opamp output and non-inverter input) would cause \$R_{IN}\$ to increases (for \$A_O\$ from 0 to 1, and for \$A_O\$ large the 1 we would create a negative resistance). But this is a different story.