N46 (39-4, UK) Determine all positive integer pairs $(a, b)$ such that $a b^{2}+b+7$ divides $a^{2} b+a+b$.

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From the given conditions, we have

i.e.,
Therefore,
$$
\begin{array}{c}
a b^{2}+b+7 \mid b\left(a^{2} b+a+b\right), \\
a b^{2}+b+7 \mid a\left(a b^{2}+b+7\right)+b^{2}-7 a, \\
a b^{2}+b+7 \mid b^{2}-7 a .
\end{array}
$$
(1) If $b^{2}-7 a>0$, then
$$
b^{2}-7 a<b^{2}<a b^{2}+b+7,
$$

which is a contradiction.
(2) If $b^{2}=7 a$, then $a, b$ should have the form $a=7 k^{2}, b=7 k\left(k \in \mathbf{N}^{*}\right)$. Clearly, $(a, b)=\left(7 k^{2}, 7 k\right)$ satisfies the condition.
(3) If $b^{2}<7 a$, then from
$$
a b^{2}+b+7 \leqslant 7 a-b^{2}<7 a,
$$

we know that $b^{2}<7$, so $b=1$ or 2.
When $b=1$, from
$$
\frac{a^{2} b+a+b}{a b^{2}+b+7}=\frac{a^{2}+a+1}{a+8}=a-7+\frac{57}{a+8}
$$

being a positive integer, we get $a+8 \mid 57$, thus we can obtain $a=11$ or 49. That is, $(a, b)=(11,1),(49,1)$.
When $b=2$, from
$$
\frac{a^{2} b+a+b}{a b^{2}+b+7}=\frac{2 a^{2}+a+2}{4 a+9}
$$

being a positive integer, and
$$
\begin{aligned}
2 \cdot \frac{2 a^{2}+a+2}{4 a+9} & =\frac{4 a^{2}+9 a+4-7 a}{4 a+9} \\
& =a-\frac{7 a-4}{4 a+9},
\end{aligned}
$$

so $\frac{7 a-4}{4 a+9}$ is a positive integer. It is easy to see that $\frac{7 a-4}{4 a+9}<2$, hence
$$
\frac{7 a-4}{4 a+9}=1, a=\frac{13}{3},
$$

which is a contradiction.
In summary, the solutions to this problem are $(a, b)=(11,1),(49,1)$ or $\left(7 k^{2}, 7 k\right), k \in \mathbf{N}^{*}$.