28 Remove one number from the set $1,2, \cdots, n+1$, and arrange the remaining numbers into a sequence $a_{1}, a_{2}, a_{3}, \cdots, a_{n}$, such that the $n$ numbers $\left|a_{1}-a_{2}\right|,\left|a_{2}-a_{3}\right|, \cdots,\left|a_{n}-a_{1}\right|$ are all different. For what values of $n(n \geqslant 3)$ can the above requirement be met?
28 It is easy to know that $a$ and $|a|$ have the same parity. Also, because
$$
\left(a_{1}-a_{2}\right)+\left(a_{2}-a_{3}\right)+\cdots+\left(a_{n}-a_{1}\right)=0,
$$

thus
$$
\left|a_{1}-a_{2}\right|+\left|a_{2}-a_{3}\right|+\cdots+a_{n}-a_{1} \mid
$$

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 $n$ distinct natural numbers, and none of them exceed $n$. Therefore, these $n$ numbers are $1,2, \cdots, n$. Hence
$$
1+2+\cdots+n=\frac{1}{2} n(n+1)
$$

must be even. So $n=4 k$ or $n=4 k-1 .(k \in \mathbf{N})$.
When $n=4 k$, remove $k+1$. Take $a_{1}=4 k+1, a_{2}=1, \cdots, a_{2 k}=k, a_{2 k-1}=3 k+$ $1, a_{2 k+2}=k+2, a_{2 k+3}=3 k, a_{2 k-4}=k+3, \cdots, a_{\{k-1}=2 k+2, a_{4 k}=$ $2 k+1$.
When $n=4 k-1$, remove $3 k$. Take $a_{1}=4 k, a_{2}=1, a_{3}=4 k-1, a_{1}=2, \cdots, a_{2 k}$
$$
\begin{array}{l}
=k, a_{2 k+1}=3 k+1, a_{2 k-2}=k+1 \cdot a_{2 k-3}=3 k-1, \cdots, a_{4 k}=2 k-1, a_{4 k-1} \\
=2 k .
\end{array}
$$

It is easy to see that both arrangements satisfy the problem's requirements.