II. (Total 50 points) Let $x_{1}, x_{2}, \cdots, x_{n}$ and $a_{1}, a_{2}, \cdots, a_{n}$ be two sets of real numbers satisfying the following conditions:
(1) $\sum_{i=1}^{n} x_{i}=0$;
(2) $\sum_{i=1}^{n}\left|x_{i}\right|=1$;
(3) $a_{1} \geqslant a_{2} \geqslant \cdots \geqslant a_{n}\left(a_{1}>a_{n}, n \geqslant 2\right)$.
Find the range of values for $m$ such that the inequality $\left| \sum_{i=1}^{n} a_{i} x_{i} \right| \leqslant m\left(a_{1}-a_{n}\right)$ always holds.
Let $n=2$, we have
\[
\left\{\begin{array}{l}
x_{1}+x_{2}=0, \\
\left|x_{1}\right|+\left|x_{2}\right|=1 .
\end{array}\right.
\]
Then $x_{1}=\frac{1}{2}, x_{2}=-\frac{1}{2}$ or $x_{1}=-\frac{1}{2}, x_{2}=\frac{1}{2}$.
From $\left|\sum_{i=1}^{n} a_{i} x_{1}\right| \leqslant m\left(a_{1}-a_{n}\right)$,
we get $\frac{1}{2}\left(a_{1}-a_{2}\right) \leqslant m\left(a_{1}-a_{2}\right) \cdot\left(a_{1}>a_{2}\right)$.
Thus, $m \geqslant \frac{1}{2}$.
Next, we prove that when $m \geqslant \frac{1}{2}$, equation (1) always holds.
\[
\begin{array}{l}
\because a_{1} \geqslant a_{i} \geqslant a_{n}, a_{1}>a_{n}(i=1,2, \cdots, n), \\
\therefore\left|\left(a_{1}-a_{i}\right)-\left(a_{i}-a_{n}\right)\right| \\
\quad \leqslant\left|\left(a_{1}-a_{1}\right)+\left(a_{i}-a_{n}\right)\right|, \\
\therefore\left|2 a_{i}-a_{1}-a_{n}\right| \leqslant\left|a_{1}-a_{n}\right|(i=1,2, \cdots, n) .
\end{array}
\]

From conditions (1) and (2), we get
\[
\begin{array}{l}
\left|\sum_{i=1}^{n} a_{i} x_{i}\right|=\left|\sum_{i=1}^{n} a_{i} x_{i}-\frac{1}{2} a_{1} \sum_{i=1}^{n} x_{i}-\frac{1}{2} a_{n} \sum_{i=1}^{n} x_{i}\right| \\
=\frac{1}{2}\left|\sum_{i=1}^{n}\left(2 a_{i}-a_{1}-a_{n}\right) x_{i}\right| \\
\leqslant \frac{1}{2} \sum_{i=1}^{n}\left|\left(2 a_{i}-a_{1}-a_{n}\right) x_{i}\right| \\
\leqslant \frac{1}{2}\left(a_{1}-a_{n}\right) \sum_{i=1}^{n}\left|x_{i}\right| \\
=\frac{1}{2}\left(a_{1}-a_{n}\right) \leqslant m\left(a_{1}-a_{n}\right) .
\end{array}
\]

Therefore, $m \geqslant \frac{1}{2}$.