2. 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 $\sum_{1 \leqslant i<j \leqslant n} x_{i} x_{j}\left(x_{i}+x_{j}\right)$. (32nd IMO Shortlist Problem)
2. Let $X=\left\{\left(x_{1}, x_{2}, \cdots, x_{n}\right) \mid x_{i} \geqslant 0, i=1,2, \cdots, n\right.$, and $\left.\sum_{i=1}^{n} x_{i}=1\right\}$, then
$$F(v)=\sum_{1 \leqslant i<j \leqslant n} x_{i} x_{j}\left(x_{i}+x_{j}\right), v=\left(x_{1}, x_{2}, \cdots, x_{n}\right) \in X$$

Obviously, when $n=2$, $\max F(v)=\frac{1}{4}$ (only when $x_{1}=x_{2}=\frac{1}{2}$), consider the case $n \geqslant 3$. Take any $v \in X$.

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 $w=\left(x_{1}, x_{2}, \cdots, x_{n-1}+x_{n}, 0\right)$.
$$\begin{aligned}
F(w)= & \sum_{1 \leqslant i<j \leqslant n-2} x_{i} x_{j}\left(x_{i}+x_{j}\right)+\sum_{i=1}^{n} x_{i}\left(x_{n-1}+x_{n}\right)\left(x_{i}+x_{n-1}+x_{n}\right)= \\
& \sum_{1 \leqslant i<j \leqslant n-2} x_{i} x_{j}\left(x_{i}+x_{j}\right)+\sum_{i=1}^{n-2} x_{i} x_{n-1}\left(x_{i}+x_{n-1}\right)+ \\
& \sum_{i=1}^{n-2} x_{i} x_{n}\left(x_{i}+x_{n}\right)+2 x_{n-1} x_{n} \sum_{i=1}^{n-2} x_{i}= \\
& F(v)+x_{n-1} x_{n}\left(2 \sum_{i=1}^{n-2} x_{i}-x_{n-1}-x_{n}\right)
\end{aligned}$$

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 $F(w) \geqslant F(v)$.
By induction, for any $v \in X$, there exists $u=(a, b, 0, \cdots, 0) \in X$, such that $F(u) \geqslant F(v)$.
Therefore, by the case $n=2$, the maximum value sought is $\frac{1}{4}$, and it is only achieved when $x_{1}=x_{2}=\frac{1}{2}, x_{3}=$ $x_{4}=\cdots=x_{n}=0$.