Taiwan IMO Selection Camp, 2022

The only integer with property is $n = 2$.
As $b \equiv -a^2 - 3 \pmod{a^2 + b + 3}$, the numerator of the given fraction satisfies
$$ab + 3b + 8 \equiv a(-a^2 - 3) + 3(-a^2 - 3) + 8 \equiv -(a + 1)^3 \pmod{a^2 + b + 3}$$
As $a^2 + b + 3$ is not divisible by $p^3$ for any prime $p$, if $a^2 + b + 3$ divides $(a+1)^3$
then it does also divide $(a+1)^2$. Since
$$0 < (a+1)^2 < 2(a^2 + b + 3),$$
we conclude $(a+1)^2 = a^2 + b + 3$. This yields $b = 2(a-1)$ and $n = 2$. The choice $(a,b) = (2,2)$ with $a^2 + b + 4 = 9$ shows that $n = 2$ indeed is a solution.