Example 7 At a banquet, $n$ guests are invited, and they sit around a round table, with seats already marked by the host with numbers $1,2,3, \cdots, n$, a total of $n$ numbers. A waiter serves the guests according to a peculiar rule: after serving one guest, he moves counterclockwise a number of seats equal to the number on the card of the seat of the guest he just served, and then serves the guest at the newly arrived seat. Similarly, he moves counterclockwise a number of seats equal to the number on the card of the seat of the guest he just served.

Find all positive integers $n \geqslant 2$, such that the host can appropriately place these $n$ cards, and the waiter can properly choose a guest to start, and according to the above rule, serve each guest.
(2003 Italian Olympiad Problem)
If $n=2 k+1\left(k \in \mathbf{N}_{+}\right)$, and the cards can be placed as required, assume the waiter serves the guests at seats $x_{1}, x_{2}$, $\cdots, x_{2 k+1}$ in sequence, then $x_{2 k+1}=2 k+1$. Otherwise, after the waiter serves the guest corresponding to card number $2 k+1$, the next guest to be served is still this guest, so $x_{1}+x_{2}+\cdots+x_{2 k}=1+2+3+\cdots+$ $2 k=k(2 k+1)$ is a multiple of $2 k+1$, which means that after the waiter serves the guest at seat number $2 k$, they return to the starting place (seat with card number $x_{1}$). Therefore, the waiter can never serve the guest corresponding to the $2 k+1$ card.

When $n=2 k\left(k \in \mathbf{N}_{+}\right)$, the host places the $2 k$ card first in a clockwise direction. Then place the other even numbers in increasing order, and finally place all the odd numbers in increasing order. The waiter serves the guest corresponding to the 1st card first, the next one to be served is the $2 k-2$nd. Then $3,2 k-4,5,2 k-6, \cdots$, $2 k-3,2,2 k-1,2 k$, satisfying the conditions of the problem.
In summary, the required positive integer $n \geqslant 2$ is all positive even numbers.