The resistors shouldn't be necessary. Literally hundreds of millions of devices work perfectly with ICs directly connected. I blame the power supply.
Pentium100 makes an important point: especially digital ICs should be properly decoupled. If you don't, switching may cause negative spikes on the power supply, which may cause false triggers in flip-flops. Use 100nF capacitors as close as possible between Vdd and ground.
Then this fishy ground. You shouldn't have two grounds! Ground of the STK500 and your circuit should be connected, why aren't they?
What you're talking about is not huge risk for a hobby project.. a few things to think about.
It's not really the \$V_F\$ at the operating current that matters, it's the \$V_F\$ at the maximum 'off' brightness that is acceptable. That voltage might be as low as 1.4-1.5V volt for a low-current red LED- in a dark room they can be quite visible with microamperes of current. Driving the output to the 3.3V (nominal level) gets us 3.3V on the output. A fresh alkaline cell with minimal load might be 1.63V/cell at room temperature (just measured one), so 3 would be 4.89V. That leaves you with 1.59V across the LED + resistor (nominal, not allowing that the 3.3V might be a few percent low).
That's way too much to ensure it's not emitting a whole bunch of light.
So, we tri-state it- that allows the output to go maybe a bit above Vcc without much current flowing. 300mV is safe, the datasheet says 500mV absolute maximum. At 500mV, we'd have 1.09V across the LED, probably enough to ensure it's off, at least under nominal conditions. The 'absolute maximum' figure is never a good thing to design to, but usually there's a caveat on this particular figure that allows that voltage or a bit more if the current is limited.
So, I do think this will work (with tri-state, not push-pull), and I also think it's acceptable enough for a hobby project assuming nobody is going to be using a battery eliminator on the circuit in the future*. Keep in mind the margin on the red LEDs is minimal and consider eschewing red in favor of yellow or orange. Or, simply add a silicon diode in series with the red LEDs (one diode can be used for several LEDs).
- If the ESD protection network in the ATMEGA328P does begin to conduct it will tend to raise up the 3.3V supply, out of control of the regulator. This is not a good thing for stability and could conceivably damage something, though the ATMEGA328P-M itself is rated for 5V operation.
I once did something like this in a commercial product (to drive a series string of LEDs with high Vf using a 5V constant-current output), but I designed a power supply with just the correct oddball voltage and appropriate temperature coefficient to match the LEDs and thus optimize the situation. I think the supply was around 8-9V. Worked a treat, easily from -20°C to 80°C (spec was 0-50°C).
Best Answer
Based on your comment, in the first case pin 1 and pin 2 are actually on different ICs. Adding a resistor to ground just ensures the lead will be a logic 0 on startup. Usually, I/O pins are initialized by default as high-impedance inputs, and don't become outputs until the microcontroller writes to the I/O configuration registers. Without the resistor, the lead would initially be floating. 4.99K or 10K are typical resistor values. If the initial state of the lead is not important, the resistor can be omitted.
The second case is really the same -- a resistor to ground ensures the EN lead will be a logic 0 on startup. If the sense of the EN lead is reversed, e.g. \$\mathsf{\small \overline{\text{EN}}}\$, then the resistor would be tied to V\$_{CC}\$ instead (same value). No decoupling capacitor is needed -- those are only for power supply pins (V\$_{CC}\$ or V\$_{DD}\$) of logic chips and the microcontroller.
The clock (SCL) and data (SDA) leads of an I²C bus each have to have a pull-up resistor to V\$_{CC}\$. This is because the I²C outputs of a master or slave device have open-drain outputs -- they only sink current (logic 0), and never source it (for logic 1). This is what allows several devices to be placed on the same bus. Again no decoupling capacitor is needed (wouldn't hurt anything, but it's not standard practice.)