19. Let $x_{i} \geqslant 0, i=1,2, \cdots, n$, and $\sum_{i=1}^{n} x_{i}=1, n \geqslant 2$, find the maximum value of $\sum_{1 \leqslant i \leqslant j \leqslant n} x_{i} x_{j}\left(x_{i}+x_{j}\right)$.
19. Let $\mathrm{X}=\left\{\left(x_{1}, x_{2}, \cdots, x_{n}\right) \mid x_{i} \geqslant 0\right.$, and $\left.\sum_{i=1}^{n} x_{i}=1\right\}$,
$$
F(v)=\sum_{1 \leqslant i<j \leqslant n} x_{i} x_{j}\left(x_{i}+x_{j}\right), v=\left(x_{1}, x_{2}, \cdots, x_{n}\right) \in \mathrm{X} .
$$

Clearly, when $n=2$, $\max F(v)=\frac{1}{4}$ (if and only if $x_{1}=x_{2}=\frac{1}{2}$). Consider the case when $n \geqslant 3$, for any $v \in Z$, let $v=\left(x_{1}, x_{2}, \cdots, x_{n}\right)$, without loss of generality, assume $x_{1} \geqslant x_{2} \geqslant \cdots \geqslant x_{n-1} \geqslant x_{n}$, take $\omega$ $=\left(x_{1}, x_{2}, \cdots, x_{n-1}+x_{n}, 0\right)$, then
$$
\begin{array}{l}
F(\omega)=\sum_{1 \leqslant i<j \leqslant n-2} x_{i} x_{j}\left(x_{i}+x_{j}\right)+\sum_{i=1}^{n-2} x_{i}\left(x_{n-1}+x_{n}\right)\left(x_{i}+x_{n-1}+x_{n}\right) \\
=F(v)+x_{n-1} x_{n}\left(2 \sum_{i=1}^{n-2} x_{i}-x_{n-1}-x_{n}\right) .
\end{array}
$$

Since $n \geqslant 3$, thus $\sum_{i=1}^{n-2} x_{i} \geqslant \frac{1}{3}, x_{n-1}+x_{n} \leqslant \frac{2}{3}$, so $F(\omega) \geqslant F(v)$. By induction, for any $v \in \mathrm{X}$, there exists $u=(a, b, \cdots, 0) \in \mathrm{X}$, such that $F(u) \geqslant F(v)$, hence, by the case when $n=2$, the maximum value sought is $\frac{1}{4}$, and it is achieved when $x_{1}=x_{2}=\frac{1}{2}, x_{3}=\cdots=x_{n}=0$.