jmc

If $n^2-19n+99=x^2$ for some positive integer $x$, then rearranging we get $n^2-19n+99-x^2=0$. Now from the quadratic formula, $n=\frac{19\pm \sqrt{4x^2-35}}{2}$ Because $n$ is an integer, this means $4x^2-35=q^2$ for some nonnegative integer $q$. Rearranging gives $(2x+q)(2x-q)=35$. Thus $(2x+q,2x-q)=(35,1)$ or $(7,5)$, giving $x=3$ or $9$. This gives $n=1,9,10,$ or $18$, and the sum is $1+9+10+18=\boxed{38}$.