In theory, for an op-amp in inverting design, the voltage gain is \$\dfrac{R_{f}}{R_{in}}\$.
For example: for a \$gain=10\$ we can use \$R_{f}=100\ \mathrm{k\Omega}\$ and \$R_{in} =10\ \mathrm{k\Omega}\$. The same gain can be obtained from \$R_{f}=1\ \mathrm{k\Omega}\$ and \$R_{in}=100\ \mathrm{\Omega}\$. What is the difference and which value is best suited?
A config like this:
Answer
If you want your op-amp to perform better at high frequencies you'll use lower value resistors to set the gain. Leakage capacitance around the feedback area might be in the order of 1pF due to circuit tracks and pads for components and this has an effect when resistor values are high.
If your feedback resistor were 100k ohm and your leakage capacitance were 1 pF, you'd find that at a frequency of: -
\$\dfrac{1}{2\pi RC} = \dfrac{1}{2\pi\times 100,000\times 1\times 10^{-12}} = 1.59MHz\$
The gain would be 3dB down on the dc gain i.e. if your dc gain is 10, at 1.59MHz the gain would be 7.07. If you need a flat response up to 32MHz then the biggest feedback resistor you can use is 5k ohm.
Op-amp data sheets are the best place to look to see what they recommend.
Taking the resistor values lower is fine but you will approach a point where the feedback resistor is starting to load the output circuits of your op-amp and you may get reduced amplitude swings and/or distortion.
But the bigger problem would be on the input resistor. To maintain a gain of ten with a feedback resistor of (say) 100 ohms means the input resistor is 10 ohms and this input resistor is the input impedance of your circuit. This would be seen by many circuits or signals as "too low" and can cause the inputted signals to distort or reduce in amplitude.
Typically you wouldn't go below 50 ohms for \$R_{IN}\$ and this means your feedback resistor is 500 ohms.
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