Answer: D

Small Hint:

Set $\dfrac{n}{20 - n} = k^2$ and solve for $n$ in terms of $k$

Big Hint:

$n = \dfrac{20k^2}{k^2 + 1}$; since $k^2$ and $k^2 + 1$ are coprime, $20$ is divisible by $k^2 + 1$

Solution:

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.