smc

Let $x^2=\frac{n}{20-n}$, with $x\ge 0$ (note that the solutions $x<0$ do not give any additional solutions for $n$). Then rewriting, $n=\frac{20x^2}{x^2+1}$. Since $\text{gcd}(x^2,x^2+1)=1$, it follows that $x^2+1$ divides $20$. Listing the factors of $20$, we find that $x=0,1,2,3$ are the only $\boxed{4}$ solutions (respectively yielding $n=0,10,16,18$).