Solution:

Work modulo $n + 2,$ where $n \equiv -2.$ Then $$3(n + 3)(n^2 + 9) \equiv 3 \cdot 1 \cdot (4 + 9) = 39 \pmod{n + 2},$$ so $n + 2$ divides $3(n+3)(n^2+9)$ exactly when $n + 2$ divides $39.$

The divisors of $39$ that are at least $3$ are $3,$ $13,$ and $39,$ giving $n = 1,$ $11,$ and $37.$ The sum is $1 + 11 + 37 = 49.$