$8 \cdot 165$ A merry-go-round has a total of $n$ seats, and boy $A$ rides for $n$ rounds. After each round, he gets up from his seat and moves to another seat in a clockwise direction, but he never completes a full circle in any single move. The number of seats (including the one he was originally sitting on) that boy $A$ passes through during a move is called the length of that move. Assuming the lengths of his $n-1$ moves are all different, find all possible values of $n$ that allow $A$ to sit in all $n$ seats over the $n$ rounds.
[Solution] (1) When $n$ is odd, the sum of the lengths of $n-1$ transfers is
$$
1+2+\cdots+(n-1)=\frac{1}{2} n(n-1) \text {. }
$$

This is a multiple of $n$. Thus, it indicates that the position after the $n$-th rotation is the same as the 1st. This means that at least one of the $n$ seats has not been sat on. Therefore, odd $n$ do not meet the requirement.
(2) When $n$ is even, the sum of the $n-1$ transfer lengths represented by $\circledast$ is no longer a multiple of $n$, so the issue in (1) no longer exists. At this time, the $n-1$ transfer lengths of $A$ can be taken as follows:
When $n=4 k+2$, they are taken sequentially as
$$
1, n-2,3, n-4, \cdots, 2 k+1,2 k, 2 k+3,2 k-2, \cdots, n-3,2, n-1 \text {. }
$$

It is easy to calculate that the sum of the first $m$ terms of the above sequence modulo $n$ is sequentially
$$
\begin{array}{l}
1, n-1,2, n-2, \cdots, k+1, n-(k+1), k+2, n-(k+2), \cdots, 2 k, \\
n-2 k, 2 k+1 .
\end{array}
$$

This precisely indicates that the $n-1$ seats occupied after $n-1$ transfers, plus the initial seat, are all different, totaling $n$ seats.
When $n=4 k$, they are taken sequentially as
$$
1, n-2,3, n-4, \cdots, 2 k-1,2 k, 2 k+1,2 k-2, \cdots, n-3,2, n-1 \text {. }
$$

Similarly, it can be verified that this method meets the requirements of the problem.
In summary, all possible values of $n$ are all even numbers.