$=6$ integer solutions.



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Solution: Let the two integer roots of the equation be $x_{1}$ and $x_{2}$. Then
$$
\left\{\begin{array}{l}
x_{1}+x_{2}=-(10 a+b), \\
x_{1} x_{2}=10 b+a .
\end{array}\right.
$$

We have $10 x_{1}+10 x_{2}+x_{1} x_{2}=-99 a(1 \leqslant a \leqslant 9)$.
Thus $\left(x_{1}+10\right)\left(x_{2}+10\right)=100-99 a$.
By the problem, $x_{1}<0, x_{2}<0$, and
$$
\overline{a b}=-\left(x_{1}+x_{2}\right), \overline{b a}=x_{1} x_{2}=99 \text {. }
$$

Let $\left|x_{1}\right| \leqslant\left|x_{2}\right|$, by $\left|x_{1}\right| \cdot\left|x_{2}\right| \leqslant 99$ we know $\left|x_{1}\right| \leqslant 9$.
Obviously, $x_{1} \neq 0$, then
$$
9 \geqslant x_{1}+10 \geqslant 1 \text {. }
$$

Discussion: (1) When $a=1$.
$$
\left(x_{1}+10\right)\left(x_{2}+10\right)=1 \text {. }
$$

By (2), only
$$
\left\{\begin{array}{l}
x_{1}+10=1, \\
x_{2}+10=1 .
\end{array}\right.
$$

That is, $x_{1}=x_{2}=-9$.
Substituting into equation (1) gives $\overline{a b}=18, \overline{b a}=81$.
(2) When $a=2$,
$$
\left(x_{1}+10\right)\left(x_{2}+10\right)=-98 \text {. }
$$

By (2) we have,
$$
x_{2}+10<0 \text {. }
$$

Thus $-98=k \cdot\left(-\frac{98}{k}\right) \cdot\left(k, \frac{98}{k}\right.$ are integers, $\left.1 \leqslant k \leqslant 9\right)$
$$
\begin{array}{l}
\therefore\left(x_{1}+10\right)\left(x_{2}+10\right) \\
=1 \times(-98)=2 \times(-49)=7 \times(-14) .
\end{array}
$$

Solving gives $\left\{\begin{array}{l}x_{1}=-9, \\ x_{2}=-108 ;\end{array}\left\{\begin{array}{l}x_{1}=-8, \\ x_{2}=-59 ;\end{array}\left\{\left\{\begin{array}{l}x_{1}=-3, \\ x_{2}=-24\end{array}\right.\right.\right.\right.$.
Substituting into equation (1) gives $\overline{a b}=27, \overline{b a}=72$.
At this time, $x_{1}=-3, x_{2}=-24$.
(3) When $a=3$,
$$
\left(x_{1}+10\right)\left(x_{2}+10\right)=-197=1 \times(-197) \text {. }
$$

By (2) we know there is no solution.
(4) When $a=4$,
$$
\begin{array}{l}
\left(x_{1}+10\right)\left(x_{2}+10\right) \\
=-296=1 \times(-296)=2 \times(-148) \\
=4 \times(-74)=8 \times(-37) .
\end{array}
$$

By (2) we know $k=1,2,4$, there is no solution.
Solving gives $x_{1}=-2, x_{2}=-47$.
Substituting into equation (1) gives $\overline{a b}=49, \overline{b a}=94$.
(5) When $a=5$,
$\left(x_{1}+10\right)\left(x_{2}+10\right)=-395=5 \times(-75)$, there is no solution.
(6) When $a=6$,
$$
\left(x_{1}+10\right)\left(x_{2}+10\right)=-494=2 \times(-247) \text {, no solution. }
$$

(7) When $a=7$,
$\left(x_{1}+10\right)\left(x_{2}+10\right)=-593=1 \times(-593)$, there is no solution.
(8) When $a=8$,
$\left(x_{1}+10\right)\left(x_{2}+10\right)=-692=4 \times(-173)$, there is no solution.
(9) When $a=9$,
$\left(x_{1}+10\right)\left(x_{2}+10\right)=-791=7 \times(-113)$, there is no solution.
In summary, when $\overline{a b}=18,27,49$, the integer solutions of the equation are: $\left\{\begin{array}{l}x_{1}=-9, \\ x_{2}=-9 ;\end{array}\left\{\begin{array}{l}x_{1}=-3, \\ x_{2}=-24 ;\end{array}\left\{\begin{array}{l}x_{1}=-2, \\ x_{2}=-47 .\end{array}\right.\right.\right.$