10. Let the natural number $n \geqslant 3$, and the real numbers $x_{1}, x_{2}, \cdots, x_{n}$ satisfy $x_{1}+x_{2}+\cdots+x_{n}=n, x_{1}^{2}+x_{2}^{2}+\cdots+x_{n}^{2}=$ $n^{2}$,
find the minimum value of $\sum_{i=1}^{n} x_{i}^{3}$ and the $\left(x_{1}, x_{2}, \cdots, x_{n}\right)$ when the minimum value is achieved.
According to the problem, $\left\{\begin{array}{l}x_{2}+x_{3}+\cdots+x_{n}=n-x_{1}, \\ x_{2}^{2}+x_{3}^{2}+\cdots+x_{n}^{2}=n^{2}-x_{1}^{2} .\end{array}\right.$ Substituting into $(n-1)\left(x_{2}^{2}+x_{3}^{2}+\cdots+x_{n}^{2}\right)$
$$
\geqslant\left(x_{2}+x_{3}+\cdots+x_{n}\right)^{2} \Rightarrow(n-1)\left(n^{2}-x_{1}^{2}\right) \geqslant\left(n-x_{1}\right)^{2} \Rightarrow x_{1}^{2}-2 x_{1}-n(n-2) \leqslant 0
$$
$\Rightarrow 2-n \leqslant x_{1} \leqslant n$, similarly, we can prove $2-n \leqslant x_{i} \leqslant n(i=2,3, \cdots, n)$.
Construct the inequality $\left(x_{i}+n-2\right)\left(x_{i}-2\right)^{2} \geqslant 0 \Rightarrow x_{i}^{3}+(n-6) x_{i}^{2}-4(n-3) x_{i}+4(n-2)$ $\geqslant 0 \Rightarrow x_{i}^{3} \geqslant-(n-6) x_{i}^{2}+4(n-3) x_{i}-4(n-2)$.
Thus, $\sum_{i=1}^{n} x_{i}^{3} \geqslant-n^{2}(n-6)+4 n(n-3)-4 n(n-2)=-n^{3}+6 n^{2}-4 n$.
Equality holds when $x_{i}=2-n$ or 2. Clearly, they cannot all be 2, nor can there be more than 1 $2-n$, only 1 $2-n$ and $n-1$ 2s. At this time, $\sum_{i=1}^{n} x_{i}=2-n+2(n-1)=n, \sum_{i=1}^{n} x_{i}^{2}=$ $(2-n)^{2}+4(n-1)=n^{2}$, which satisfies the problem.