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.