Example 2 Let real numbers $x_{1}, x_{2}, \cdots, x_{1997}$ satisfy
(1) $-\frac{1}{\sqrt{3}} \leqslant x_{i} \leqslant \sqrt{3}(i=1,2, \cdots, 1997)$;
(2) $x_{1}+x_{2}+\cdots+x_{1997}=-318 \sqrt{3}$.

Try to find: $x_{1}^{12}+x_{2}^{12}+\cdots+x_{1997}^{12}$'s maximum value, and explain the reason.
Given that $f(x)=x^{12}$ is a convex function, then $\sum_{i=1}^{197} x_{i}^{12}$ achieves its maximum value when at least 1996 of $x_{1}, x_{2}, \cdots, x_{1997}$ are equal to $-\frac{1}{\sqrt{3}}$ or $\sqrt{3}$.

Assume that there are $t$ instances of $-\frac{1}{\sqrt{3}}$ and $1996-t$ instances of $\sqrt{3}$. Then the remaining one is
$$
\begin{array}{l}
-318 \sqrt{3}-\left[-\frac{1}{\sqrt{3}} t+(1996-t) \sqrt{3}\right] \\
=\frac{4}{\sqrt{3}} t-2314 \sqrt{3} .
\end{array}
$$

From the given information,
$$
\begin{array}{l}
-\frac{1}{\sqrt{3}} \leqslant \frac{4}{\sqrt{3}} t-2314 \sqrt{3} \leqslant \sqrt{3} \\
\Rightarrow 1735.25 \leqslant t \leqslant 1736.25 .
\end{array}
$$

Noting that $t \in \mathbf{N}$, hence $t=1736$.
Thus, $1996-t=260$.
Therefore, according to the proposition, when $x_{1}, x_{2}, \cdots, x_{1997}$ have 1736 instances of $-\frac{1}{\sqrt{3}}$, 260 instances of $\sqrt{3}$, and one instance of $\frac{2 \sqrt{3}}{3}$, $\sum_{i=1}^{1997} x_{i}^{12}$ achieves its maximum value, which is
$$
\begin{array}{l}
\left(-\frac{1}{\sqrt{3}}\right)^{12} \times 1736+(\sqrt{3})^{12} \times 260+\left(\frac{2 \sqrt{3}}{3}\right)^{12} \\
=189548 .
\end{array}
$$