4.4.9 ** Let $x_{1}, x_{2}, \cdots, x_{n}$ and $a_{1}, a_{2}, \cdots, a_{n}(n>2)$ be two sets of arbitrary real numbers satisfying the conditions:
(1) $x_{1}+x_{2}+\cdots+x_{n}=0$;
(2) $\left|x_{1}\right|+\left|x_{2}\right|+\cdots+\left|x_{n}\right|=1$;
(3) $a_{1} \geqslant a_{2} \geqslant \cdots \geqslant a_{n}$. To make the inequality $\mid a_{1} x_{1}+a_{2} x_{2}+\cdots+ a_{n} x_{n} \mid \leqslant A\left(a_{1}-a_{n}\right)$ hold, find the minimum value of the number $A$.
Let $i_{1}, i_{2}, \cdots, i_{s}, i_{1}, j_{2}, \cdots, j_{t}$ be a permutation of $1,2, \cdots, n$, and such that
$$
x_{i_{1}} \geqslant x_{i_{2}} \geqslant \cdots \geqslant x_{i_{s}} \geqslant 0 > x_{j_{1}} \geqslant x_{j_{2}} \geqslant \cdots \geqslant x_{j_{t}}.
$$

Let $\lambda = x_{i_{1}} + x_{i_{2}} + \cdots + x_{i_{s}}$ and $\mu = - (x_{j_{1}} + x_{j_{2}} + \cdots + x_{j_{t}})$. According to the problem, we have
$$
\begin{array}{c}
x_{1} + x_{2} + \cdots + x_{n} = (x_{i_{1}} + \cdots + x_{i_{s}}) + (x_{j_{1}} + \cdots + x_{j_{2}}) = \lambda - \mu = 0, \\
|x_{1}| + |x_{2}| + \cdots + |x_{n}| = (x_{i_{1}} + \cdots + x_{i_{s}}) - (x_{j_{1}} + \cdots + x_{j_{4}}) = \lambda + \mu = 1,
\end{array}
$$

so $\lambda = \mu = \frac{1}{2}$.
Without loss of generality, assume $a_{1} x_{1} + a_{2} x_{2} + \cdots + a_{n} x_{n} \geqslant 0$ (otherwise, if $a_{1} x_{1} + a_{2} x_{2} + \cdots + a_{n} x_{n} < 0$, take $x_{i}^{\prime} = -x_{i}, i = 1,2, \cdots, n$, then $x_{1}^{\prime}, x_{2}^{\prime}, \cdots, x_{n}^{\prime}$ still satisfy (1) and (2), and
$$
|a_{1} x_{1}^{\prime} + a_{2} x_{2}^{\prime} + \cdots + a_{n} x_{n}^{\prime}| = |a_{1} x_{1} + a_{2} x_{2} + \cdots + a_{n} x_{n}|,
$$

which does not affect the generality of the conclusion), by the rearrangement inequality, we have
$$
\begin{aligned}
& a_{1} x_{1} + a_{2} x_{2} + \cdots + a_{n} x_{n} \\
\leqslant & a_{1} x_{i_{1}} + a_{2} x_{i_{2}} + \cdots + a_{s} x_{i_{s}} + a_{s+1} x_{j_{1}} + \cdots + a_{n} x_{j_{t}} \\
\leqslant & a_{1} (x_{i_{1}} + x_{i_{2}} + \cdots + x_{i_{s}}) + a_{n} (x_{j_{1}} + \cdots + x_{j_{i}}) \\
= & \frac{1}{2} (a_{1} - a_{n}).
\end{aligned}
$$

On the other hand, if we take $x_{1} = \frac{1}{2}, x_{2} = x_{3} = \cdots = x_{n-1} = 0, x_{n} = -\frac{1}{2}$, then $x_{1}, x_{2}, \cdots, x_{n}$ satisfy (1) and (2), and
$$
|a_{1} x_{1} + a_{2} x_{2} + \cdots + a_{n} x_{n}| = \frac{1}{2} (a_{1} - a_{n}).
$$

Therefore, the smallest value of $A$ that satisfies the given inequality is $\frac{1}{2}$.