Example 8 (40th IMO National Team Selection Test) For non-negative real numbers $x_{1}, x_{2}, \cdots, x_{n}$ satisfying the condition $\sum_{i=1}^{n} x_{i}=1$, find the maximum value of $\sum_{i=1}^{n}\left(x_{j}^{4}-x_{j}^{5}\right)$.
Prove (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 $x_{i}+x_{j}+x_{k} \leqslant 1$.

By (1), we can merge these two numbers into one number and add a 0, making the value of the sum in the problem larger. After a finite number of adjustments, we are left with two non-zero numbers. 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., $x y=\frac{1}{6}$, $\left(x^{4}-x^{5}+y^{4}-y^{5}\right)_{\text {max }}=\frac{1}{12}$.
This is the maximum value sought in the problem. The non-zero $x_{i}$ that achieve this maximum value are only two, denoted by $x, y$, then $x, y>0, x+y=1, x y=\frac{1}{6}$.

Solving for $x=\frac{3+\sqrt{3}}{6}, y=\frac{3-\sqrt{3}}{6}$ or $x=\frac{3-\sqrt{3}}{6}, y=\frac{3+\sqrt{3}}{6}$.
Thus, when $x_{1}, x_{2}, \cdots, x_{n}$ have only these two non-zero numbers, $\left[\sum_{j=1}^{n}\left(x_{j}^{4}-x_{j}^{5}\right)\right]_{\text {max }}=\frac{1}{12}$.