Find all positive integers $n$ such that $n+2$ divides $n^{3}+3 n+29$.

 untranslated text has been directly translated while preserving the formatting and structure.
Necessary Condition (NC): Let $n$ be suitable. We have $n+2$ divides $(n+2) n^{2}$ as well as the difference $(n+2) n^{2}-\left(n^{3}+3 n+29\right)$. Thus, $n+2 \mid 2 n^{2}-3 n-29$. Similarly, $n+2 \mid 2 n(n+2)-\left(2 n^{2}-3 n-29\right)=7 n+29$ and then $n+2 \mid 7 n+29-7(n+2)=15$.

Therefore, considering that $n \geq 2, n+2$ is 3, 5, or 15. So $n$ is 1, 3, or 13.

Sufficient Condition (SC):

- For $n=1, n+2=3$ and $n^{3}+3 n+29=33$. We have $3 \mid 33$
- For $n=3, n+2=5$ and $n^{3}+3 n+29=65$. We have $5 \mid 65$
- For $n=13, n+2=15$ and $n^{3}+3 n+29=2265$. We have $3 \mid 2265$ and $5 \mid 2265$ which means $15 \mid 2265$

The solutions to the problem are therefore 1, 3, and 13.