Every of $n$ guests invited to a dinner has got an invitation denoted by a number from $1$ to $n$. The guests will be sitting around a round table with $n$ seats. The waiter has decided to derve them according to the following rule. At first, he selects one guest and serves him/her at any place. Thereafter, he selects the guests one by one: having chosen a guest, he goes around the table for the number of seats equal to the preceeding guest's invitation number (starting from the seat of the preceeding guest), and serves the guest there.
Find all $n$ for which he can select the guests in such an order to serve all the guests.
1. **Restate the problem**: We need to determine for which values of \( n \) it is possible to serve all guests around a round table following a specific rule. The rule is that after serving a guest, the waiter moves a number of seats equal to the invitation number of the last served guest to serve the next guest.

2. **Initial placement and movement**: We start by placing an arbitrary number in the first seat. From there, we count a number of spaces equal to the number just placed (wrapping around if necessary) to determine the next seat to be filled. 

3. **Example for \( n = 4 \)**:
   - Place 2 in the first seat: \(|2| \_ \_ \_\)
   - Move 2 spaces: \(2 \_ \| \_ \_\)
   - Place 3 in the next seat: \(2 \_ |3| \_\)
   - Move 3 spaces (wrapping around): \(2 \| \_ 3 \_\)
   - Place 1 in the next seat: \(2 |1| 3 \_\)
   - Move 1 space: \(2 1 |3| \_\)
   - The next seat is already filled, so this arrangement fails.

4. **Valid arrangement for \( n = 4 \)**:
   - Place 3 in the first seat: \(|3| \_ \_ \_\)
   - Move 3 spaces: \(3 \_ \_ \| \_\)
   - Place 1 in the next seat: \(3 \_ \_ |1|\)
   - Move 1 space: \(3 \| \_ 1 \_\)
   - Place 4 in the next seat: \(3 \_ |4| 1\)
   - Move 4 spaces (wrapping around): \(3 \_ 4 |1|\)
   - Place 2 in the next seat: \(3 |2| 4 1\)
   - This arrangement is valid.

5. **General case for even \( n \)**:
   - For even \( n \), we can use the sequence: \((n-1, n-3, n-5, \ldots, 3, 1, n, n-2, n-4, \ldots, 4, 2)\).
   - This sequence ensures that each number is placed in a unique seat and the last seat remains empty until the final placement of \( n \).

6. **Sum of numbers for odd \( n \)**:
   - The sum of the numbers from 1 to \( n-1 \) is \(\frac{(n-1)n}{2}\).
   - For odd \( n \), this sum is divisible by \( n \), meaning after \( n-1 \) placements, we return to the first seat.
   - Since the first seat is already filled, it is impossible to place \( n \) in the last step.

7. **Conclusion**:
   - For even \( n \), it is possible to serve all guests following the given rule.
   - For odd \( n \), it is not possible to serve all guests following the given rule.

The final answer is for all even \( \boxed{ n } \).