Find all integers $n \in \mathbb{N}$ such that $n \mid\left(n^{2}+2 n+27\right)$.
Suppose that $n$ is a solution. Since $n \mid\left(n^{2}+2 n\right)$, then $n \mid\left(\left(n^{2}+2 n+27\right)-\left(n^{2}+2 n\right)\right)$, i.e., $n \mid 27$. Conversely, we can show similarly that if $n \mid 27$, then $n \mid\left(n^{2}+2 n+27\right)$.

Thus, the solutions are the divisors of 27, i.e., $1, 3, 9$, and 27.