6・117 Let real numbers $x_{1}, x_{2}, \cdots, x_{1997}$ satisfy the following two conditions:
(1) $-\frac{1}{\sqrt{3}} \leqslant x_{i} \leqslant \sqrt{3} \quad(i=1,2, \cdots, 1997)$;
(2) $x_{1}+x_{2}+\cdots+x_{1997}=-318 \sqrt{3}$.

Find: $x_{1}^{12}+x_{2}^{12}+\cdots+x_{1997}^{12}$'s maximum value, and explain the reason.
[Solution] For any set of $x_{1}, x_{2}, \cdots, x_{1997}$ that satisfies the given conditions, if there exist such $x_{i}$ and $x_{j}$:
$$
\sqrt{3}>x_{i} \geqslant x_{j}>-\frac{1}{\sqrt{3}} .
$$

We denote
$$
\begin{array}{l}
x_{i}+x_{j}=2 m, \\
x_{i}-x_{j}=2 h .
\end{array}
$$

Then
$$
\begin{array}{l} 
x_{i}=m+h, x_{j}=m-h, \\
x_{i}^{12}+x_{j}^{12}=(m+h)^{12}+(m-h)^{12} \\
=2 \sum_{0 \leq k=2 l \leq 12} C_{12}^{k} m^{12-k} h^{k} .
\end{array}
$$

Under the condition that the sum $x_{i}+x_{j}=2 m$ remains constant, $x_{i}^{12}+x_{j}^{12}$ increases as $h>0$ increases. We agree to take
$$
h=\min \left\{\sqrt{3}-m, m-\left(-\frac{1}{\sqrt{3}}\right)\right\},
$$

then $x_{i}^{12}+x_{j}^{12}$ reaches its maximum value.
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 rest of the variables take values of $\sqrt{3}$ or $-\frac{1}{\sqrt{3}}$.

Suppose $u$ variables take the value $-\frac{1}{\sqrt{3}}, v$ variables take the value $\sqrt{3}, w$ variables take a value in $\left(-\frac{1}{\sqrt{3}}, \sqrt{3}\right) \cdot$ Clearly,
$$
w=0 \text { or } 1 \text {. }
$$

When $w=1$, let the value of this variable be $t$. Thus, we have
$$
\left\{\begin{array}{l}
u+v+w=1997, \\
-\frac{1}{\sqrt{3}} \cdot u+\sqrt{3} v+t w=-318 \sqrt{3} .
\end{array}\right.
$$
(2) $\times \sqrt{3}+$ (1) gives:
$$
4 v+(\sqrt{3} t+1) w=1043 \text {. }
$$

Since $(\sqrt{3} t+1) w=1043-4 v$ is an integer, and
$$
0 \leqslant(\sqrt{3} t+1) w<4,
$$

it follows that $(\sqrt{3} t+1) w$ is the remainder when 1043 is divided by 4. Therefore,
$$
w=1, t=\frac{2}{\sqrt{3}} \text {. }
$$

Substituting into (1) and (2), we solve to get
$$
u=1736, v=260 \text {. }
$$

Therefore, the maximum value of $x_{1}^{12}+x_{2}^{12}+\cdots+x_{1997}^{12}$ is
$$
\begin{aligned}
& \left(-\frac{1}{\sqrt{3}}\right)^{12} \cdot u+(\sqrt{3})^{12} \cdot v+t^{12} \\
= & \frac{1736+4096}{729}+729 \times 260 \\
= & 189548
\end{aligned}
$$