Example 4. In the acute triangle $\triangle A B C$, the lengths of the three sides $a, b, c$ are all integers, and $a>b>c, a+b+c=20$. Find $\angle A + \angle C$.
First, determine the range of the maximum side length $a$.

From $b+c>a>b>c$, we get $20=a+b+c>2a$, i.e., $a<\frac{20}{2}=10$ and $a>\frac{20}{3}$.
$\therefore \frac{20}{3}<a<10$.
(1) When $a=7$, $b+c=13$, then only $b=8, c=5$.

At this time, $\cos A=\frac{b^{2}+c^{2}-a^{2}}{2bc}=\frac{1}{2}>0$, so $\triangle ABC$ is an acute triangle.

Also, $\cos B=\frac{1}{2}$, i.e., $\angle B=60^{\circ}$, then $\angle A+\angle C=120^{\circ}$.
(2) When $a=8$, $b+c=12$, then only $b=7, c=5$.

At this time, $\cos A=\frac{b^{2}+c^{2}-a^{2}}{2bc}=\frac{1}{7}>0$, so $\triangle ABC$ is an acute triangle.

Also, $\cos B=\frac{1}{2}$, i.e., $\angle B=60^{\circ}$, then $\angle A+\angle C=120^{\circ}$.
(3) When $a=9$, $b+c=11$, then there are $b=8, c=3; b=7, c=4; b=6, c=5$. At this time, $\triangle ABC$ is not an acute triangle in all cases.
In summary, $\angle A+\angle C=120^{\circ}$.