(4) Given an integer $n \geqslant 3$, real numbers $a_{1}, a_{2}, \cdots, a_{n}$ satisfy $\min _{1 \leqslant i<j \leqslant n} \mid a_{i}-$ $a_{j} \mid=1$. Find the minimum value of $\sum_{k=1}^{n}\left|a_{k}\right|^{3}$.
Let's assume $a_{1}<a_{2}<\cdots<a_{n}$, then for $1 \leqslant k \leqslant n$, we have
$$
\left|a_{k}\right|+\left|a_{n-k+1}\right| \geqslant\left|a_{n-k+1}-a_{k}\right| \geqslant|n+1-2 k|
$$

Therefore, $\sum_{k=1}^{n}\left|a_{k}\right|^{3}=\frac{1}{2} \sum_{k=1}^{n}\left(\left|a_{k}\right|^{3}+\left|a_{n+1-k}\right|^{3}\right)=$
$$
\begin{array}{l}
\frac{1}{2} \sum_{k=1}^{n}\left(\left|a_{k}\right|+\left|a_{n+1-k}\right|\right)\left(\frac { 3 } { 4 } \left(\left|a_{k}\right|-\right.\right. \\
\left.\left.\left|a_{n+1-k}\right|\right)^{2}+\frac{1}{4}\left(\left|a_{k}\right|+\left|a_{n+1-k}\right|\right)^{2}\right) \geqslant \\
\frac{1}{8} \sum_{k=1}^{n}\left(\left|a_{k}\right|+\left|a_{n+1-k}\right|\right)^{3} \geqslant \\
\frac{1}{8} \sum_{k=1}^{n}|n+1-2 k|^{3}
\end{array}
$$

When $n$ is odd,
$$
\sum_{k=1}^{n}|n+1-2 k|^{3}=2 \cdot 2^{3} \cdot \sum_{i=1}^{\frac{n-1}{2}} i^{3}=\frac{1}{4}\left(n^{2}-1\right)^{2}
$$

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, $\sum_{k=1}^{n}\left|a_{k}\right|^{3} \geqslant \frac{1}{32}\left(n^{2}-1\right)^{2}$, and when $n$ is even, $\sum_{k=1}^{n}\left|a_{k}\right|^{3} \geqslant \frac{1}{32} n^{2}\left(n^{2}-2\right)$, with equality holding 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 $n$).