4. Given an integer $n(n \geqslant 3)$, real numbers $a_{1}, a_{2}, \cdots, a_{n}$ satisfy $\min _{1<i<j<n}\left|a_{i}-a_{j}\right|=1$. Find the minimum value of $\sum_{k=1}^{n}\left|a_{k}\right|^{3}$.
(Zhu Huawei)
4. Let $a_{1}<a_{2}<\cdots<a_{n}$.

Then for $1 \leqslant k \leqslant n$ we have
$$
\begin{array}{l}
\left|a_{k}\right|+\left|a_{n-k+1}\right| \geqslant\left|a_{n-k+1}-a_{k}\right| \\
\geqslant|n+1-2 k| . \\
\text { Hence } \sum_{k=1}^{n}\left|a_{k}\right|^{3}=\frac{1}{2} \sum_{k=1}^{n}\left(\left|a_{k}\right|^{3}+\left|a_{n+1-k}\right|^{3}\right) \\
=\frac{1}{2} \sum_{k=1}^{n}\left(\left|a_{k}\right|+\left|a_{n+1-k}\right|\right) . \\
\quad\left[\frac{3}{4}\left(\left|a_{k}\right|-\left|a_{n+1-k}\right|\right)^{2}+\frac{1}{4}\left(\left|a_{k}\right|+\left|a_{n+1-k}\right|\right)^{2}\right] \\
\geqslant \frac{1}{8} \sum_{k=1}^{n}\left(\left|a_{k}\right|+\left|a_{n+1-k}\right|\right)^{3} \\
\left.\geqslant \frac{1}{8} \sum_{k=1}^{n} \right\rvert\, n+1-2 k^{3}
\end{array}
$$

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

Therefore, when $n$ is odd,
$$
\sum_{k=1}^{n}\left|a_{k}\right|^{3} \geqslant \frac{1}{32}\left(n^{2}-1\right)^{2} \text {; }
$$

When $n$ is even,
$$
\sum_{k=1}^{n}\left|a_{k}\right|^{3} \geqslant \frac{1}{32} n^{2}\left(n^{2}-2\right) \text {. }
$$

Equality holds when $a_{i}=i-\frac{n+1}{2}(i=1,2, \cdots, n)$.

Thus, the minimum value of $\sum_{k=1}^{n}\left|a_{k}\right|^{3}$ is $\frac{1}{32}\left(n^{2}-1\right)^{2}$ (for odd $n$), or $\frac{1}{32} n^{2}\left(n^{2}-2\right)$ (for even $n$).