AMC 12 Mock Exam 2 Problem 10

Since $n^2+12=(n+12)(n-12)+156$, the number $n+12$ divides $n^2+12$ exactly when it divides $156=2^2\cdot3\cdot13$. The divisors of $156$ that exceed $12$ are $13,26,39,52,78,156$, giving $n=1,14,27,40,66,144$. Their sum is $\boxed{292}$.