What is the importance of the \$50\Omega\$ resistors in this RF op-amp circuit?
I think you're confusing an aerial and transmission line when you mention wavelength in the order of 1 meter - the transmission line equation is essentially independent of the frequency (and hence wavelength).

You're correct to think in terms of transmission lines . The other thing to realize is that this op amp is a current feedback amp (and not the 'normal' voltage feedback amp). 

{see http://en.wikipedia.org/wiki/Current-feedback_operational_amplifier }

The 49.9 ohm terminating the first op amp output sets its output impedance to 50 ohm (nominal). The second 49.9 ohm resistor terminates what is in effect a non resonant  50 ohm transmission line and produces a flatly tuned circuit. The result of this termination is to reduce the gain of the first stage by half which looks rather odd but is necessary to maintain the flat tuning of the stage.



Getting back to the 39pf capacitor. It boosts the signal at the high frequency end and compensates for the roll-off in gain but will mis-match the termination leading to some reflection at high frequency.

The circuit is designed to slot straight into a 50 ohm system transmission line system.