Solve using the adjustment method to explore the maximum value of the sum in the problem.
(1) First, for $x, y>0$, we compare $(x+y)^{4}-(x+y)^{5}+0^{4}-0^{5}$ with $x^{4}-x^{5}+y^{4}-y^{5}$:
$$\begin{aligned} & (x+y)^{4}-(x+y)^{5}+0^{4}-0^{5}-\left(x^{4}-x^{5}+y^{4}-y^{5}\right) \\ = & x y\left(4 x^{2}+6 x y+4 y^{2}\right)-x y\left(5 x^{3}+10 x^{2} y+10 x y^{2}+y^{3}\right) \\ \geqslant & \frac{7}{2} x y\left(x^{2}+2 x y+y^{2}\right)-5 x y\left(x^{3}+3 x^{2} y+3 x y^{2}+y^{3}\right) \\ = & \frac{1}{2} x y(x+y)^{2}[7-10(x+y)] . \end{aligned}$$

As long as $x, y>0, x+y>0$, then among them, there must be two whose sum is $0 \leqslant \frac{2}{3}$. For this case, let $x, y>0, x+y=1$. For this scenario,
$$\begin{aligned} & x^{4}-x^{5}+y^{4}-y^{5}=x^{4}(1-x)+y^{4}(1-y)=x y\left(x^{3}+y^{3}\right) \\ = & x y\left[(x+y)^{3}-3 x y(x+y)\right]=x y(1-3 x y) \\ = & \frac{1}{3}(3 x y)(1-3 x y) . \end{aligned}$$

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 required by the problem. The $x_{1}, x_{2}, \cdots, x_{n}$ that can achieve the maximum value, only two of them are not equal to 0. Let $x$ and $y$ represent these two numbers, then $x, y>0, x+y=1, x y=\frac{1}{6}$.

Solving the quadratic equation $\lambda^{2}-\lambda+\frac{1}{6}=0$ gives $x=\frac{3+\sqrt{3}}{6}, y=\frac{3-\sqrt{3}}{6}$, or 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 $\frac{1}{12}$.