2002 AMC 10B Problem 16

Suppose $\dfrac{n}{20 - n} = k^2$ for some integer $k \ge 0.$ Solving, $$n = \dfrac{20k^2}{k^2 + 1}.$$ Since $k^2$ and $k^2 + 1$ share no common factor, $k^2 + 1$ must divide $20.$ This happens only for $k = 0,$ $1,$ $2,$ $3,$ giving $k^2 + 1 = 1,$ $2,$ $5,$ $10.$ The corresponding values $n = 0,$ $10,$ $16,$ $18$ are all integers, so there are $4$ such $n.$ Thus, the correct answer is D.