China Mathematical Competition

We may assume $0 \le x \le y \le z \le 1$. Then
$$M = \sqrt{y-x} + \sqrt{z-y} + \sqrt{z-x}.$$
Since
$$\sqrt{y-x} + \sqrt{z-y} \le \sqrt{2[(y-x)+(z-y)]} = \sqrt{2(z-x)},$$
we have
$$M \le \sqrt{2(z-x)} + \sqrt{z-x} = (\sqrt{2} + 1) \sqrt{z-x} \le \sqrt{2} + 1.$$
The equality holds if and only if $y-x = z-y$, $x = 0$, $z = 1$ (i.e. $x = 0$, $y = \frac{1}{2}$, $z = 1$).
Therefore, the answer is $M_{\max} = \sqrt{2} + 1$.