9. Find all positive integers $a, b$, such that
$$\left[\frac{a^{2}}{b}\right]+\left[\frac{b^{2}}{a}\right]=\left[\frac{a^{2}+b^{2}}{a b}\right]+a b$$
9. When $a=b$, the left side $=a+b$, and the right side $=ab+2$. Also, $ab+2-a-b=(a-1)(b-1)+1>0$, so the left and right sides are not equal, hence $a \neq b$.

Assume without loss of generality that $ab>a$, thus $a^2-b2a$, which is a contradiction. Therefore, $b<a^2+2$.

In summary, $b=a^2+1$. Direct verification shows that $b=a^2+1$ meets the requirements. Therefore, the solutions $(a, b)=\left(m, m^2+1\right)$ or $\left(m^2+1, m\right), m \in \mathbf{N}^{*}$.