One. (Total 20 points) $a$ is a three-digit number, $b$ is a one-digit number, and $\frac{a}{b} 、 \frac{a^{2}+b^{2}}{a b+1}$ are both integers. Find the maximum and minimum values of $a+b$.
Given $b \mid a$, and $(a b+1) \mid\left(a^{2}+b^{2}\right)$, we have
$$
\begin{array}{l}
\frac{a^{2}+b^{2}}{a b+1}=\frac{a}{b}+\frac{b^{2}-\frac{a}{b}}{a b+1} . \\
\because a b+1>a b>b^{2}>b^{2}-\frac{a}{b}>-1-a b,
\end{array}
$$
$\therefore b^{2}-\frac{a}{b}=0$, which means $a=b^{3}$.
When $b=5$, $a=125$; when $b=9$, $a=729$. Thus, $(a+b)_{\text {max }}=738,(a+b)_{\text {min }}=130$.