II. (40 points) Let integer $n \geqslant 3$, real numbers $x_{1}, x_{2}, \cdots, x_{n}$ satisfy $\min _{1 \leqslant i<j \leqslant n}\left|x_{i}-x_{j}\right|=1$. Find $\min _{x_{1}, x_{2}, \cdots, x_{n} \in \mathbf{R}} \sum_{k=1}^{n}\left|x_{k}\right|^{3}$.
Let's assume $x_{1}<x_{2}<\cdots<x_{n}$.
From $\min _{1 \leqslant i<j \leqslant n}\left|x_{i}-x_{j}\right|=1$, we know that for $1 \leqslant k \leqslant n$, we have
$$
\begin{array}{l}
\left|x_{k}\right|+\left|x_{n-k+1}\right| \geqslant\left|x_{n-k+1}-x_{k}\right| \\
\geqslant\left|\sum_{j=k}^{n-k}\left(x_{j+1}-x_{j}\right)\right| \geqslant|n+1-2 k| . \\
\text { Then } \sum_{k=1}^{n}\left|x_{k}\right|^{3}=\sum_{k=1}^{n} \frac{\left|x_{k}\right|^{3}+\left|x_{n+1-k}\right|^{3}}{2} \\
\geqslant \sum_{k=1}^{n}\left(\frac{\left|x_{k}\right|+\left|x_{n+1-k}\right|}{2}\right)^{3} \\
\geqslant \frac{1}{8} \sum_{k=1}^{n}|n+1-2 k|^{3} .
\end{array}
$$

When $n$ is odd,
$$
\begin{array}{l}
\sum_{k=1}^{n}|n+1-2 k|^{3} \\
=2 \times 2^{3} \sum_{i=1}^{\frac{n-1}{2}} i^{3}=\frac{1}{4}\left(n^{2}-1\right)^{2} .
\end{array}
$$

When $n$ is even,
$$
\begin{array}{l}
\sum_{k=1}^{n}|n+1-2 k|^{3}=2 \sum_{i=1}^{\frac{n}{2}}(2 i-1)^{3} \\
=2\left(\sum_{j=1}^{n} j^{3}-\sum_{i=1}^{\frac{n}{2}}(2 i)^{3}\right)=\frac{1}{4} n^{2}\left(n^{2}-2\right) .
\end{array}
$$

The equality holds in both cases when $x_{i}=i-\frac{n+1}{2}(i=1,2, \cdots, n)$.
Therefore, $\min _{x_{1}, x_{2}, \cdots, x_{n} \in \mathbf{R}} \sum_{k=1}^{n}\left|x_{k}\right|^{3}$
$=\left\{\begin{array}{ll}\frac{1}{32}\left(n^{2}-1\right)^{2}, & n \text { is odd; } \\ \frac{1}{32} n^{2}\left(n^{2}-2\right), & n \text { is even. }\end{array}\right.$