7 For non-negative real numbers $x_{i}(i=1,2, \cdots, n)$ satisfying $x_{1}+x_{2}+\cdots+x_{n}=1$, find the maximum value of $\sum_{j=1}^{n}\left(x_{j}^{4}-x_{j}^{5}\right)$.
7. When $n=1$, $\sum_{j=1}^{n}\left(x_{j}^{4}-x_{j}^{5}\right)=0$.

When $n=2$, $\sum_{j=1}^{n}\left(x_{j}^{4}-x_{j}^{5}\right)=\left(x_{1}^{4}+x_{2}^{4}\right)-\left(x_{1}+x_{2}\right)\left(x_{1}^{4}-x_{1}^{3} x_{2}+x_{1}^{2} x_{2}^{2}-\right.$ $\left.x_{1} x_{2}^{3}+x_{2}^{4}\right)=x_{1}^{3} x_{2}-x_{1}^{2} x_{2}^{2}+x_{1} x_{2}^{3}=x_{1} x_{2}\left(1-3 x_{1} x_{2}\right)$.

Since $x_{1} x_{2} \leqslant \frac{1}{4}$, the maximum value is $\frac{1}{12}$.
When $n \geqslant 3$, then
$$\begin{aligned}
& -\left(x^{4}+y^{4}-x^{5}-y^{5}\right)+\left[(x+y)^{4}-(x+y)^{5}\right] \\
= & {\left[(x+y)^{4}-x^{4}-y^{4}\right]-\left[(x+y)^{5}-x^{5}-y^{5}\right] } \\
= & x y\left(4 x^{2}+4 y^{2}+6 x y\right)-x y\left(5 x^{3}+5 y^{3}+10 x y^{2}+10 x^{2} y\right) \\
\geqslant & x y\left(4 x^{2}+4 y^{2}+6 x y\right)-5 x y(x+y)^{3} \\
= & x y\left(\frac{7}{2} x^{2}+\frac{7}{2} y^{2}+\frac{x^{2}+y^{2}}{2}+6 x y\right)-5 x y(x+y)^{3} \\
\geqslant & x y\left(\frac{7}{2} x^{2}+\frac{7}{2} y^{2}+7 x y\right)-5 x y(x+y)^{3}>0
\end{aligned}$$

This is equivalent to $\frac{7}{2} x y(x+y)^{2} \cdot[7-10(x+y)]>0$, i.e., $x+y<\frac{7}{10}$.
When $n \geqslant 3$, there are always 2 numbers whose sum $<\frac{2}{3}<\frac{7}{10}$. By continuously replacing $x$ and $y$ with $x+y$, it can eventually be reduced to the case when $n=2$.

In summary, the maximum value of $\sum_{j=1}^{n}\left(x_{j}^{4}-x_{j}^{5}\right)$ is $\frac{1}{12}$.