AIME II

Solution:
Note that $n \equiv -2 \pmod{(n+2)}$ and thus $3(n+3)(n^2+9) \equiv 3 \cdot 1 \cdot 13 = 39 \pmod{(n+2)}$. Therefore $n$ is a positive solution if and only if $n+2$ is a divisor of $39$ that is at least $3$. Those divisors are $3$, $13$, and $39$, and the corresponding values of $n$ are $1$, $11$, and $37$, respectively. The requested sum is $1 + 11 + 37 = 49$.

OR
Note that $\gcd(n+2, n+3) = 1$, so $n+2$ must divide the product $3(n^2+9)$. This product can be rewritten as
$$3(n^2 + 9) = 3(n^2 - 4 + 13) = 3(n + 2)(n - 2) + 39.$$
Therefore $n+2$ must be a divisor of $39$ that is at least $3$, as in the first solution.