$a_{n}$, such that $\left|a_{1}-a_{2}\right|,\left|a_{2}-a_{3}\right|, \cdots,\left|a_{n}-a_{1}\right|$ are all distinct, for what kind of number $n(n \geqslant 3)$ can the above requirement be met?
Since $a$ and $|a|$ have the same parity, and
$$
\left(a_{1}-a_{2}\right)+\left(a_{2}-a_{3}\right)+\cdots+\left(a_{n}-a_{1}\right)=0,
$$

it follows that $\left|a_{1}-a_{2}\right|+\left|a_{2}-a_{3}\right|+\cdots+\left|a_{n}-a_{1}\right|$ is even.
On the other hand, $\left|a_{1}-a_{2}\right|,\left|a_{2}-a_{3}\right|, \cdots,\left|a_{n}-a_{1}\right|$ are distinct and do not exceed $n$, so these $n$ numbers are $1,2, \cdots, n$. Therefore, $1+2+\cdots+n=\frac{1}{2} n(n+1)$ is even, so $n=4 k$ or $n=4 k-1$ (where $k$ is a positive integer).

When $n=4 k$, remove $k+1$, and take $a_{1}=4 k+1, a_{2}=1, a_{3}=4 k, a_{4}=2, \cdots, a_{2 k}=$ $k, a_{2 k+1}=3 k+1, a_{2 k+2}=k+2, a_{2 k+3}=3 k, a_{2 k+4}=k+3, \cdots, a_{4 k-1}=2 k+2$, $a_{4 k}=2 k+1$.
When $n=4 k-1$, remove $3 k$, and take $a_{1}=4 k, a_{2}=1, a_{3}=4 k-1, a_{4}=2, \cdots$,
$$
\begin{array}{l}
a_{2 k}=k, a_{2 k+1}=3 k-1, a_{2 k+2}=k+1, a_{2 k+3}=3 k-2, \cdots, a_{4 k-2}=2 k-1, \\
a_{1 k-1}=2 k .
\end{array}
$$

In summary, $n=4 k$ or $4 k-1$ (where $k$ is a positive integer).