AMC 12 Mock Exam 4 Problem 13

Suppose $n^2+30n=m^2$ with $m\ge0$. Then $(n+15)^2-m^2=225$, so $(n+15-m)(n+15+m)=225$ with $0<n+15-m\le n+15+m$. The factor pairs of $225$ are $(1,225),(3,75),(5,45),(9,25),(15,15)$, giving $n+15=\frac{d+e}{2}=113,39,25,17,15$, i.e., $n=98,24,10,2,0$. Excluding $n=0$, the sum is $98+24+10+2=\boxed{134}$.