Note that

${n^2 \over n+4}={n^2-16 \over n+4}+{16 \over n+4}={(n+4)(n-4) \over n+4}+{16 \over n+4}=n-4+{16 \over n+4}$.

So when $n> 12$, the remainder when $n^2$ is divided by $n+4$ is always $16$.

For $1 \leq n \leq 12$, the remainder when $n^2$ is divided by $n+4$ is shown in the table below.

$n\quad\quad\quad$ | 1 | 2 | 2 | 4 | 5 | 6 | 7 | 8 | 9 | 10 | 11 | 12 |

$n+4$ | 5 | 6 | 7 | 8 | 9 | 10 | 11 | 12 | 13 | 14 | 15 | 16 |

remainder | 1 | 4 | 2 | 0 | 7 | 6 | 5 | 4 | 3 | 2 | 1 | 0 |

So there are $9$ different remainders, namely $0, 1, 2, 3, 4, 5, 6, 7, 16$.

*This problem is taken from the UKMT Mathematical Challenges.*