A boy goes $n$ times at a merry-go-round with $n$ seats. After every time he moves in the clockwise direction and takes another seat, not making a full circle. The number of seats he passes by at each move is called the length of the move. For which $n$ can he sit at every seat, if the lengths of all the $n-1$ moves he makes have different lengths? 
[i]V. New[/i]
To determine for which \( n \) the boy can sit at every seat on a merry-go-round with \( n \) seats, given that the lengths of all \( n-1 \) moves he makes are different, we need to analyze the problem for both even and odd \( n \).

1. **Case 1: \( n = 2k \) (even number of seats)**

   - We use the sequence of moves: \( 2k-2, 1, 2k-4, 3, \ldots, 2k-3, 0 \).
   - To understand why this works, note that moving \( m \) seats clockwise is equivalent to moving \( 2k-2-m \) seats counterclockwise.
   - This sequence alternates between large and small moves, ensuring that each move is unique and covers all seats without repeating any seat until all seats are visited.
   - The sequence effectively builds on the ends of a growing block of places where the boy has sat before, eventually covering all seats.

2. **Case 2: \( n = 2k+1 \) (odd number of seats)**

   - We consider the sequence of moves: \( 1, 2, \ldots, 2k \).
   - The sum of these lengths is:
     \[
     \sum_{i=1}^{2k} i = \frac{(2k+1)(2k)}{2} = k(2k+1)
     \]
   - This sum is a multiple of \( 2k+1 \), meaning that after \( 2k \) moves, the boy will return to the initial seat, leaving one seat unvisited.
   - Therefore, it is impossible for the boy to sit at every seat if \( n \) is odd.

Conclusion:
- For \( n = 2k \) (even), the boy can sit at every seat.
- For \( n = 2k+1 \) (odd), the boy cannot sit at every seat.

The final answer is \( \boxed{ n } \) must be even.