21. (1999 National Team Selection Test) For non-negative real numbers $x_{1}, x_{2}$, $\cdots, x_{n}$ satisfying the condition $x_{1}+x_{2}+\cdots+x_{n}=1$, find the maximum value of $\sum_{j=1}^{n}\left(x_{j}^{4}-x_{j}^{5}\right)$.
21. Use the adjustment method to explore the maximum value of the sum in the problem.

First, for $x, y>0$, we compare $(x+y)^{4}-(x+y)^{5}+0^{4}-0^{5}$ with $x^{4}-x^{5}+y^{4}-y^{5}$,
$$
\begin{aligned}
& (x+y)^{4}-(x+y)^{5}+0^{4}-0^{5}-\left(x^{4}-x^{5}+y^{4}-y^{5}\right) \\
= & x y\left(4 x^{2}+6 x y+4 y^{2}\right)-x y\left(5 x^{3}+10 x^{2} y+10 x y^{2}+5 y^{3}\right) \\
\geqslant & \frac{7}{2} x y\left(x^{2}+2 x y+y^{2}\right)-5 x y\left(x^{3}+3 x^{2} y+3 x y^{2}+y^{3}\right) \\
= & \frac{1}{2} x y(x+y)^{2}[7-10(x+y)] .
\end{aligned}
$$

As long as $x, y>0, x+y>0$, then, among them, there must be two whose sum $\leqslant \frac{2}{3}0, x+y=1$.

For this case, $x^{4}-x^{3}+y^{4}-y^{6}=x^{4}(1-x)+y^{4}(1-y)=x y\left(x^{3}+y^{3}\right)=x y\left[(x+y)^{3}-3 . x y(x+\right.$ $y)]=x y(1-3 x y)=\frac{1}{3}(3 x y)(1-3 x y)$.
When $3 x y=\frac{1}{2}$, the above expression reaches its maximum value, $x^{4}-x^{5}+y^{4}-y^{5}=\frac{1}{6}\left(1-\frac{1}{2}\right)=\frac{1}{12}$.
This is the maximum value sought in the problem, and the $x_{1}, x_{2}, \cdots, x_{n}$ that can achieve this maximum value are such that only two are not equal to 0, represented by $x$ and $y$, then $x, y>0, x+y=1, x y=\frac{1}{6}$.
Solving the quadratic equation $\lambda^{2}-\lambda+\frac{1}{6}=0$ gives $x=\frac{3+\sqrt{3}}{6}, y=\frac{3-\sqrt{3}}{6}$.
Of course, it can also be $x=\frac{3-\sqrt{3}}{6}, y=\frac{3+\sqrt{3}}{6}$.
Verification shows that if $x_{1}, x_{2}, \cdots, x_{n}$ have only these two non-zero numbers, then the sum in the problem indeed reaches the maximum value $\frac{1}{12}$.