Example 3 Let real numbers $x_{1}, x_{2}, \cdots, x_{1997}$ satisfy the following two conditions:
(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 the maximum value of $x_{1}^{12}+x_{2}^{12}+\cdots+x_{197}^{12}$, and explain the reason.
Since $x_{1}+x_{2}+\cdots+x_{1997}$ is a continuous function on a closed domain, $x_{1}^{12}+x_{2}^{12}+\cdots+x_{1997}^{12}$ must have a maximum value. Fixing $x_{3}, x_{4}, \cdots, x_{1997}$, then $x_{1}+x_{2}=c$ (a constant), let $x_{1}=x$, then $x_{2}=c-x$, thus
$$x_{1}^{12}+x_{2}^{12}+\cdots+x_{1997}^{12}=x^{12}+(x-c)^{12}+A=f(x) \text {. }$$

Since $f^{\prime \prime}(x)=132 x^{10}+132(x-c)^{10}>0$, $f(x)$ is a convex function, so when $f(x)\left(-\frac{1}{\sqrt{3}} \leqslant x \leqslant \sqrt{3}\right)$ reaches its maximum, there must be $x \in\left\{-\frac{1}{\sqrt{3}}, \sqrt{3}\right\}$. By symmetry, for any two variables $x_{i} 、 x_{j}(1 \leqslant i<j \leqslant 1997)$, when $x_{1}^{12}+x_{2}^{12}+\cdots+x_{1997}^{12}$ reaches its maximum value, $x_{i} 、 x_{j}$ must have one belonging to $\left\{-\frac{1}{\sqrt{3}}, \sqrt{3}\right\}$. Therefore, when $x_{1}^{12}+x_{2}^{12}+\cdots+x_{199}^{12}$ reaches its maximum value, $x_{1}, x_{2}, \cdots, x_{1997}$ must have at least 1996 belonging to $\left\{-\frac{1}{\sqrt{3}}, \sqrt{3}\right\}$. Without loss of generality, assume that $x_{1}, x_{2}, \cdots, x_{1997}$ have $u$ being $-\frac{1}{\sqrt{3}}, v$ being $\sqrt{3}, w$ belonging to the interval $\left(-\frac{1}{\sqrt{3}}, \sqrt{3}\right)$, where $w=0$ or 1. When $w=1$, let the variable belonging to the interval $\left(-\frac{1}{\sqrt{3}}, \sqrt{3}\right)$ be $t$, then
$$u+v+w=1997,-\frac{1}{\sqrt{3}} u+\sqrt{3} v+t w=-318 \sqrt{3} .$$

Eliminating $u$, we get $4 v+(\sqrt{3} t+1) w=1043$, so $(\sqrt{3} t+1) w$ is an integer, and $(\sqrt{3} t+1) w \equiv 1043 \equiv 3(\bmod 4)$. Also, by $-\frac{1}{\sqrt{3}}<t<\sqrt{3}$, we get $0<\sqrt{3} t+1<4$. Since $w=0$ or 1, we have $0 \leqslant(\sqrt{3} t+1) w<4$. Therefore, $(\sqrt{3} t+1) w=3$, so $w \neq 0$, i.e., $w=1$, thus $t=\frac{2}{\sqrt{3}}$. Then $4 v+3=1043$, solving for $v$ gives $v=260$, so $u=1997-v-w=1736$.

In summary, the maximum value of $x_{1}^{12}+x_{2}^{12}+\cdots+x_{1997}^{12}$ is
$$\begin{aligned}
\left(-\frac{1}{\sqrt{3}}\right)^{12} u+(\sqrt{3})^{12} v+t^{12} & =\left(\frac{1}{3}\right)^{6} \times 1736+3^{6} \times 260+\left(\frac{2}{\sqrt{3}}\right)^{12} \\
& =189548 .
\end{aligned}$$