Solution:
It is possible to arrange the name-tags in the desired way if and only if $n$ is even.
If $n=2k$ is even, one possible solution is the following: proceeding clockwise along the table, the master of ceremonies first places name-tag number $2k$, then all the even ones in increasing order, and then all the odd ones in increasing order. If now the waiter starts from name-tag $1$ and follows the rule, he goes around the whole table. Indeed, one checks that after having served an odd number $i$, the waiter goes to number $2k-i-1$, which is even, and then to number $i+2$, which is again odd.

The waiter's path will therefore be as follows: $1, 2k-2, 3, 2k-4, 5, 2k-6$, and so on, ending the path by serving number $2k$.

We now show that if $n=2k+1$ is odd, then there is no arrangement of the name-tags that works. Suppose indeed that such an arrangement exists, and that the waiter completes the table by serving, in order, the numbers $x_{1}, x_{2}, \ldots, x_{2k+1}$. Then $x_{2k+1}=2k+1$, since if he were to continue after having served number $2k+1$, the waiter would have to make a complete lap of the table and thus return to the same place. But then the sum of the preceding numbers is
$$x_{1}+x_{2}+\ldots+x_{2k}=1+2+\ldots+2k=\frac{2k(2k+1)}{2}=k(2k+1)$$
which is a multiple of $2k+1$. It follows that after having served the first $2k$ guests, the waiter has returned to the starting point, and thus he will never serve number $2k+1$.