[Solution] First, note that when $a<b$,
$$|x-a|+|x-b|=\left\{\begin{array}{l} a+b-2 x, \text { when } x \leqslant a, \\ -a+b, \text { when } a \leqslant x \leqslant b, \\ 2 x-a-b, \text { when } x \geqslant b. \end{array}\right.$$

Therefore, on each point of the interval $a \leqslant x \leqslant b$, the sum $|x-a|+|x-b|$ reaches its minimum value. We will solve this problem based on this observation.

Without loss of generality, assume the numbers $a_{1}, a_{2}, \cdots, a_{n}$ form an increasing sequence, that is,
$$a_{1}<a_{2}<\cdots<a_{n}$$

When $n=2 m$ (where $m$ is a natural number), the expression
$$y=\left|x-a_{1}\right|+\left|x-a_{2}\right|+\cdots+\left|x-a_{n}\right|$$

on the right can be divided into $m$ groups, that is,
$$\begin{aligned} y= & \left(\left|x-a_{1}\right|+\left|x+a_{n}\right|\right)+\left(\left|x-a_{2}\right|+\left|x-a_{n-1}\right|\right)+ \\ & \cdots+\left(\left|x-a_{m}\right|+\left|x-a_{m+1}\right|\right) \end{aligned}$$

The sum $y_{i}=\left|x-a_{i}\right|+\left|x-a_{n+1-i}\right| \quad(i=1,2, \cdots, m)$ is constant on the interval $a_{i} \leqslant x \leqslant a_{n+1-i}$, and this constant is its minimum value. Since each interval $a_{i} \leqslant x \leqslant a_{n+1-i}$ includes the next interval $a_{i+1} \leqslant x \leqslant a_{n+1-(i+1)}$, all intervals have a common part, which is the interval $a_{m} \leqslant x \leqslant a_{m+1}$.

On each point of the interval $a_{m} \leqslant x \leqslant a_{m+1}$, all $y_{i}$ achieve their minimum values, and thus $y$ achieves its minimum value on this interval. To calculate this value, we can set $x=a_{m}$ or $x=a_{m+1}$ in (2). This value equals
$$-a_{1}-a_{2}-\cdots-a_{n}+a_{m+1}+\cdots+a_{n}$$

When $n=2 m+1(m$ is a natural number or 0$)$, the right side of (1) can be rewritten as
$$\begin{aligned} y= & \left(\left|x-a_{1}\right|+\left|x-a_{n}\right|\right)+\left(\left|x-a_{2}\right|+\left|x-a_{n-1}\right|\right)+ \\ & \cdots+\left(\left|x-a_{m}\right|+\left|x-a_{m+2}\right|\right)+\left|x-a_{m+1}\right| \end{aligned}$$

Similar to the case when $n$ is even, it is not difficult to verify that when $x=a_{m+1}$, each $y_{i}=1 x$ $-a_{i}|+| x-a_{n+1-i} \mid$ reaches its minimum value. And at this time, $\left|x-a_{m+1}\right|=0$. Therefore, the last term in (3) also reaches its minimum value. Thus, when $x=a_{m+1}$, $y$ reaches its minimum value. According to (3), this minimum value equals $-a_{1}-a_{2}-\cdots-a_{m}+a_{m+2}+$ $a_{m+3}+\cdots+a_{n}$