XIV APMO

By the symmetry of the problem, we may suppose that $a\le b$. Notice that $b^2-a\ge 0$, so that if $\frac{a^2+b}{b^2-a}$ is a positive integer, then $a^2+b\ge b^2-a$. Rearranging this inequality and factorizing, we find that $(a+b)(a-b+1)\ge 0$. Since $a,b>0$, we must have $a\ge b-1$.

We therefore have two cases:

Case 1: $a=b$. Substituting, we have $$\frac{a^2+a}{a^2-a}=\frac{a+1}{a-1}=1+\frac{2}{a-1}$$ which is an integer if and only if $(a-1)\mid 2$. As $a>0$, the only possible values are $a-1=1$ or $2$. Hence, $(a,b)=(2,2)$ or $(3,3)$.

Case 2: $a=b-1$. Substituting, we have $$\frac{b^2+a}{a^2-b}=\frac{(a+1{)}^2+a}{a^2-(a+1)}=\frac{a^2+3a+1}{a^2-a-1}=1+\frac{4a+2}{a^2-a-1}.$$ Once again, notice that $4a+2>0$, and hence, for $\frac{4a+2}{a^2-a-1}$ to be an integer, we must have $4a+2\ge a^2-a-1$, that is, $a^2-5a-3\le 0$. Hence, since $a$ is an integer, we can bound $a$ by $1\le a\le 5$. Checking all the ordered pairs $(a,b)=(1,2),(2,3),\dots ,(5,6)$, we find that only $(1,2)$ and $(2,3)$ satisfy the given conditions.

Thus, the ordered pairs that work are $$(2,2),(3,3),(1,2),(2,3),(2,1),(3,2)$$ where the last two pairs follow by symmetry.