2002 AMC 12B Problem 12

Set $\dfrac{n}{20-n}=k^2.$ Solving, $n=\dfrac{20k^2}{k^2+1}.$ Since $k^2$ and $k^2+1$ are coprime, $k^2+1$ must divide $20,$ which happens only for $|k|=0,1,2,3.$ The signs of $k$ do not change $n;$ these give $n=0,$ $10,$ $16,$ and $18,$ which is four values. Thus, the correct answer is D.