4. In the acute triangle $\triangle A B C$, the lengths of the three sides $a, b, c$ are all integers, and $a<b<c, a+b+c=20$. Then $\angle A$ $+\angle C=$ $\qquad$ .
4. $120^{\circ}$.
$$
\because 20=(a+b)+c>c+c=2 c,
$$

there is $c<\frac{20}{3}$.
$$
\therefore \frac{20}{3}>c>0, \text{ then } \triangle A B C \text{ is an acute triangle.}
$$
$$
\begin{array}{l}
\because \cos B=\frac{c^{2}+a^{2}-b^{2}}{2 c a}=\frac{1}{2}, \\
\therefore \angle B=60^{\circ},
\end{array}
$$

then $\angle A+\angle C=120^{\circ}$.
(2) When $c=9$, from $a+b=11$ and $a<b<c$, we get $a=3, b=8 ; a=4, b=7 ; a=5, b=6$. In this case, $\cos C<0$, so $\triangle A B C$ is not an acute triangle.
Combining (1) and (2), we know that $\angle A+\angle C=120^{\circ}$.