How many positive integers $n$ exist such that $\frac{2 n^{2}+4 n+18}{3 n+3}$ is an integer?
As

$$
\frac{2 n^{2}+4 n+18}{3 n+3}=\frac{2}{3} \frac{\left(n^{2}+2 n+1\right)+8}{n+1}=\frac{1}{3}\left(2 n+2+\frac{16}{n+1}\right)
$$

it follows that $n+1$ must divide 16. Thus, $n$ must belong to the set $\{1,3,7,15\}$. In each of these cases, we have

| $n$ | $\frac{2 n^{2}+4 n+18}{3 n+3}$ |
| :---: | :---: |
| 1 | 4 |
| 3 | 4 |
| 7 | 6 |
| 15 | 11 |

Therefore, for the four values of $n$, 1, 3, 7, and 11, $\frac{2 n^{2}+4 n+18}{3 n+3}$ is an integer.