By the symmetry of the problem, we may suppose that $a \leq b$. Notice that $b^{2}-a \geq 0$, so that if $\frac{a^{2}+b}{b^{2}-a}$ is a positive integer, then $a^{2}+b \geq b^{2}-a$. Rearranging this inequality and factorizing, we find that $(a+b)(a-b+1) \geq 0$. Since $a, b>0$, we must have $a \geq b-1$. 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 $\frac{b^{2}+a}{a^{2}-b}=\frac{(a+1)^{2}+a}{a^{2}-(a+1)}=\frac{a^{2}+3 a+1}{a^{2}-a-1}=1+\frac{4 a+2}{a^{2}-a-1}$. For $\frac{4 a+2}{a^{2}-a-1}$ to be an integer, we must have $4 a+2 \geq a^{2}-a-1$, that is, $a^{2}-5 a-3 \leq 0$. Hence, since $a$ is an integer, we can bound $a$ by $1 \leq a \leq 5$. Checking all the ordered pairs $(a, b)=(1,2),(2,3), \ldots,(5,6)$, we find that only $(1,2)$ and $(2,3)$ satisfy the given conditions. Thus, the ordered pairs that work are $(2,2),(3,3),(1,2),(2,3),(2,1),(3,2)$.