We can see that $4n \ge n^2+3,$ so we need $0 \ge n^2-4n+3=(n-1)(n-3),$ note that this happens only when $n$ is in the range $[1,3].$ We can see that by testing $n=1,2,3$ the only solutions are $\boxed{n=1,3}.$

mathmax12 wrote: We can see that $4n \ge n^2+3,$ so we need $0 \ge n^2-4n+3=(n-1)(n-3),$ note that this happens only when $n$ is in the range $[1,3].$ We can see that by testing $n=1,2,3$ the only solutions are $\boxed{n=1,3}.$ What if $n=0$?

Note that $0$ works and $-n$ works iff $n$ works, so we look for positive solutions. In this case we must have $$n^2+3 \le 4n \iff (n-2)^2\le 1 \iff n \le 3.$$ Now $n = 1, 3$ are the only working positive sols, so the full solution set is $$\boxed{n \in \{-3,-1,0,1,3\}}.$$

$\frac{4n}{n^2+3}\in \mathbb{Z}\iff n^2+3|4n$ Let's do some bounding : $n^2+3\le |4n|\implies n^2-|4n|+4\le 1\implies (|n|-2)^2\le 1\implies ||n|-2|\le 1\implies -1\le |n|-2\le 1\implies 1\le |n|\le 3$ Thus : $-3\le n \le 3$ Verifying gives that : $S=\{-3,-1,0,1,3\}$