Example 1 Let real numbers $x_{1}, x_{2}, \cdots, x_{1}, \cdots 77$ 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 $x_{1}^{12}+x_{2}^{12}+\cdots+x_{1997}^{12}$. (1997 China Mathematical Olympiad Problem)
Solve for any set of $x_{1}, x_{2}, \cdots, x_{1997}$ that satisfies the given conditions. If there are such $x_{i}$ and $x_{j}$: $\sqrt{3}>x_{i} \geqslant x_{j}>-\frac{1}{\sqrt{3}}$, then let $m=\frac{1}{2}\left(x_{i}+x_{j}\right), h_{0}=\frac{1}{2}\left(x_{i}-x_{j}\right)=x_{i}-m=m-x_{j}$. We observe that
$$(m+h)^{12}+(m-h)^{12}=2 \sum_{0 \leqslant k=2 l \leqslant 12} \mathrm{C}_{12}^{k} m^{12-k} h^{k}$$

increases as $h>0$ increases. We agree to take $h=\min \left\{\sqrt{3}-m, m-\left(-\frac{1}{\sqrt{3}}\right)\right\}$, and replace $x_{i}$ and $x_{j}$ with $x^{\prime}{ }_{i}=m+h, x^{\prime}{ }_{j}=m-h$. The sum of the elements remains unchanged, and the sum of the 12th powers of the elements increases. Therefore, the maximum value of the sum of the 12th powers can only be achieved in the following scenario: at most one variable takes a value in $\left(-\frac{1}{\sqrt{3}}, \sqrt{3}\right)$, and the other variables are either $-\frac{1}{\sqrt{3}}$ or $\sqrt{3}$.

Suppose $u$ variables take the value $-\frac{1}{\sqrt{3}}, v$ variables take the value $\sqrt{3}, w(=0$ or 1$)$ variables take a value in $\left(-\frac{1}{\sqrt{3}}, \sqrt{3}\right)$ (if there is one, denote this value as $t$). Thus,
$$\left\{\begin{array}{l}
u+v+w=1997 \\
-\frac{1}{\sqrt{3}} u+\sqrt{3} v+t w=-318 \sqrt{3}
\end{array}\right.$$

From this, we get
$$4 v+(\sqrt{3} t+1) w=1043$$

Since $(\sqrt{3} t+1) w=1043-4 v$ is an integer, and $0 \leqslant 1043-4 v<4$, $(\sqrt{3} t+1) w$ is the remainder when 1043 is divided by 4. According to this, we find
$$v=260, t=\frac{2}{\sqrt{3}}, u=1736$$

According to the above discussion, the maximum value of $x_{1}^{12}+x_{2}^{12}+\cdots+x_{1997}^{12}$ is
$$\begin{array}{l}
\left(-\frac{1}{\sqrt{3}}\right)^{12} u+(\sqrt{3})^{12} v+t^{12}=\frac{1736+4096}{729}+729 \times 260= \\
8+189540=189548
\end{array}$$