One, (40 points) Let $n \in \mathbf{Z}, n \geqslant 2$, and non-negative real numbers $x_{1}, x_{2}, \cdots, x_{n}$ satisfy $x_{1}+x_{2}+\cdots+x_{n}=1$. Find
$$
\sum_{1 \leqslant i<j \leqslant n} x_{i} x_{j}\left(x_{i}+x_{j}\right)
$$

the maximum and minimum values.
Obviously, $\sum_{1 \leqslant i<j \leqslant n} x_{i} x_{j}\left(x_{i}+x_{j}\right) \geqslant 0$, when
$$
\left(x_{1}, x_{2}, x_{3}, \cdots, x_{n}\right)=(1,0,0, \cdots, 0)
$$

the equality holds.
Thus, the minimum value sought is 0.
$$
\begin{array}{l}
\text { Let } F\left(x_{1}, x_{2}, \cdots, x_{n}\right)=\sum_{1 \leqslant i<j \leqslant n} x_{i} x_{j}\left(x_{i}+x_{j}\right) \\
=\sum_{1 \leqslant i<j \leqslant n}\left(x_{i}^{2} x_{j}+x_{j}^{2} x_{i}\right)=\sum_{i=1}^{n} x_{i}^{2}\left(1-x_{i}\right) \\
=\sum_{i=1}^{n} x_{i}\left(x_{i}-x_{i}^{2}\right) .
\end{array}
$$

By the AM-GM inequality,
$x_{i}-x_{i}^{2} \leqslant\left(\frac{x_{i}+1-x_{i}}{2}\right)^{2}=\frac{1}{4}$.
Thus, $F\left(x_{1}, x_{2}, \cdots, x_{n}\right) \leqslant \frac{1}{4} \sum_{i=1}^{n} x_{i}=\frac{1}{4}$, when
$$
\left(x_{1}, x_{2}, x_{3}, x_{4}, \cdots, x_{n}\right)=\left(\frac{1}{2}, \frac{1}{2}, 0,0, \cdots, 0\right)
$$

the maximum value is achieved.
Thus, the maximum value sought is $\frac{1}{4}$.