In a $\triangle A B C$, it is given that
$$
\begin{gathered}
\quad \cos 3 A+\cos 3 B+\cos 3 C+\cos 3 \pi=0 \\
\Rightarrow \quad 2 \cos \frac{3(A+B)}{2} \cos \frac{3(A-B)}{2}+\cos 3 C-1=0 \\
\Rightarrow-2 \sin \left(\frac{3 C}{2}\right) \cos \frac{3(A-B)}{2}+1-2 \sin ^2\left(\frac{3 C}{2}\right)-1=0 \\
\Rightarrow \quad \sin \frac{3 C}{2}\left[\cos \frac{3(A-B)}{2}-\cos \frac{3(A+B)}{2}\right]=0 \\
\Rightarrow \quad \sin \frac{3 A}{2} \sin \frac{3 B}{2} \sin \frac{3 C}{2}=0 \\
\text { Means either } A=\frac{2 \pi}{3} \text { or } B=\frac{2 \pi}{3} \text { or } C=\frac{2 \pi}{3}
\end{gathered}
$$
So, the least value of the sum of two of its angles is
$$
\pi-\frac{2 \pi}{3}=\frac{\pi}{3}
$$