Fraction is an integer!

We need $n^2+7$ to divide $n^3+3$ , and hence we need $n^2 + 7$ to divide $n(n^2+7) - (n^3+3) = 7n-3$ . Since $n > 0$ . this implies that $n^2+7 \le 7n-3$ , and hence $(n-2)(n-5) \; = \; n^2 - 7n + 10 \; \le \; 0$ so that $2 \le n \le 5$ . Checking shows that $\frac{n^3+3}{n^2+7}$ is an integer precisely when $n=2,5$ , and so the answer is $2^2 + 5^2 = \boxed{29}$ .