Let $x_{k} \in[-2,2](k=1,2, \cdots, 2013)$,
and
$x_{1}+x_{2}+\cdots+x_{2013}=0$. Try to find
$$
M=x_{1}^{3}+x_{2}^{3}+\cdots+x_{2013}^{3}
$$

the maximum value.
(Liu Kangning)
Given $x_{i} \in[-2,2](i=1,2, \cdots, 2013)$, we know that $x_{i}^{3}-3 x_{i}=\left(x_{i}-2\right)\left(x_{i}+1\right)^{2}+2 \leqslant 2$.
The equality holds if and only if $x_{i}=2$ or -1.
Noting that, $\sum_{i=1}^{2013} x_{i}=0$.
Thus, $M=\sum_{i=1}^{2013} x_{i}^{3}=\sum_{i=1}^{2013}\left(x_{i}^{3}-3 x_{i}\right)$
$$
\leqslant \sum_{i=1}^{2013} 2=4026 \text {. }
$$

When $x_{1}, x_{2}, \cdots, x_{2013}$ include 671 values of 2 and 1342 values of -1, the equality holds.
Therefore, the maximum value of $M$ is 4026.