Example 8 For non-negative real numbers $x_{1}, x_{2}, \cdots, x_{n}$ satisfying $x_{1}+x_{2}+\cdots+x_{n}=1$, find the maximum value of $\sum_{j=1}^{n}\left(x_{j}^{4}-x_{j}^{5}\right)$.
(40th IMO Chinese National Team Selection Exam)
Explanation: (1) For $x, y > 0$,
$$
\begin{array}{l}
(x+y)^{4}-(x+y)^{5}+0^{4}-0^{5}-\left(x^{4}-x^{5}+y^{4}-y^{5}\right) \\
= x y\left(4 x^{2}+6 x y+4 y^{2}\right)- \\
x y\left(5 x^{3}+10 x y^{2}+10 x^{2} y+5 y^{3}\right) \\
= \frac{1}{2} x y\left(7 x^{2}+12 x y+7 y^{2}+x^{2}+y^{2}\right)- \\
5 x y\left(x^{3}+2 x y^{2}+2 x^{2} y+y^{3}\right) \\
\geqslant \frac{7}{2} x y\left(x^{2}+2 x y+y^{2}\right)- \\
5 x y\left(x^{3}+2 x y^{2}+2 x^{2} y+y^{3}\right) \\
> \frac{1}{2} x y(x+y)^{2}[7-10(x+y)] .
\end{array}
$$

Therefore, when $x > 0, y > 0, 7-10(x+y) > 0$, the above expression must be greater than 0.
(2) If there are fewer than two non-zero numbers among $x_{1}, x_{2}, \cdots, x_{n}$, then the sum in the problem is 0.

Now consider the case where there are at least two non-zero numbers among $x_{1}, x_{2}, \cdots, x_{n}$.

If there are three numbers $x_{i}, x_{j}, x_{k} > 0$, then there must be two numbers whose sum is less than or equal to $\frac{2}{3}1$, which contradicts $x_{i}+x_{j}+x_{k} \leqslant 1$.

From (1), we know that these two numbers can be combined into one number, and another number 0 can be added, making the value of the sum in the problem larger. After a finite number of adjustments, only two non-zero numbers remain. Let's assume these are $x, y > 0, x+y=1$. For this case, we have
$$
\begin{array}{l}
x^{4}-x^{5}+y^{4}-y^{5} \\
=x^{4}(1-x)+y^{4}(1-y) \\
=x^{4} y+x y^{4}=x y\left(x^{3}+y^{3}\right) \\
=x y(x+y)\left(x^{2}-x y+y^{2}\right) \\
=x y\left[(x+y)^{2}-3 x y\right] \\
=\frac{1}{3} \times 3 x y(1-3 x y) \leqslant \frac{1}{3} \times\left(\frac{1}{2}\right)^{2}=\frac{1}{12} .
\end{array}
$$

When $3 x y=1-3 x y$, i.e., $x y=\frac{1}{6}$,
$$
\left(x^{4}-x^{5}+y^{4}-y^{5}\right)_{\max }=\frac{1}{12} \text{. }
$$

This is the maximum value sought in the problem. The non-zero numbers among the $x_{i}$ are only two, denoted by $x, y$, then
$$
x, y > 0, x+y=1, x y=\frac{1}{6} \text{. }
$$

Solving this, we get $x=\frac{3+\sqrt{3}}{6}, y=\frac{3-\sqrt{3}}{6}$ or $x=\frac{3-\sqrt{3}}{6}$, $y=\frac{3+\sqrt{3}}{6}$.

Therefore, when $x_{1}, x_{2}, \cdots, x_{n}$ have only these two non-zero numbers, $\left[\sum^{n}\left(x_{j}^{4}-x_{j}^{5}\right)\right]_{\max }=\frac{1}{12}$.