Two players - Andriy and Olesya play the following game. On a table there is a round cake, which is cut by one of the players into $2n$ ($n>1$) sectors (pieces), pairwise distinct in weights. Weight of each piece is known to both players. Then they take pieces according to the following rules. At first, Olesya takes $1$ piece, then Andriy takes $2$ pieces such that the remaining pieces form a sector. Then they in turns take $2$ pieces such that after each turn the remaining pieces form a sector. In their last move, one of the players takes one last piece. Each of the players aims to get more of a cake than the opponent. For which $n$ can Olesya cut the cake such that she wins if in her first move she takes the smallest piece?

[i] Proposed by Bohdan Rublyov [/i]
To determine for which \( n \) Olesya can cut the cake such that she wins if in her first move she takes the smallest piece, we need to analyze the game strategy for different values of \( n \).

1. **Initial Setup and Simplification**:
   - The cake is cut into \( 2n \) sectors with distinct weights.
   - Olesya takes the smallest piece first.
   - Andriy then takes 2 pieces such that the remaining pieces form a sector.
   - They continue taking 2 pieces each turn, ensuring the remaining pieces form a sector.
   - The last move involves taking the final piece.

   We can simplify the problem by considering the weights of the pieces as a sequence of positive reals \( a_1, a_2, \ldots, a_{2n} \) arranged in a circle. We can subtract the minimum weight from each piece, making the smallest piece weight zero. Thus, we have \( a_1 = 0 \) and \( a_2, a_3, \ldots, a_{2n} \) are positive reals.

2. **Case Analysis for Different \( n \)**:
   - **Odd \( n \)**:
     - For \( n \) odd, Olesya can ensure her win by strategically placing a very large piece \( a_3 \) such that it is greater than half the total weight. Let \( a_3 = 10^N \) for a large \( N \), and the other pieces be negligible in comparison.
     - If Andriy takes any piece other than \( a_3 \), Olesya can take \( a_3 \) in her next move and win.
     - If Andriy takes pieces from the right end, Olesya will follow by taking the next two rightmost pieces, ensuring she eventually takes \( a_3 \).
     - This strategy works because the parity of \( n \) ensures that Olesya will have the opportunity to take \( a_3 \) before Andriy can.

   - **\( n = 2 \)**:
     - For \( n = 2 \), the game involves 4 pieces. Andriy can always take the two largest pieces in his first move, leaving Olesya with the smaller pieces. Thus, Olesya cannot win if she starts with the smallest piece.

   - **Even \( n \ge 6 \)**:
     - For even \( n \ge 6 \), Olesya can place three large pieces \( a_3 = 10^N + 2 \), \( a_{2n-3} = 10^N + 1 \), and \( a_5 = 10^N \) for a large \( N \), with the rest being negligible.
     - If Andriy takes any pair of pieces, Olesya can always respond by taking one of the large pieces and another piece, ensuring she gets a significant portion of the total weight.
     - This strategy ensures that Olesya can always secure more than half the total weight by the end of the game.

   - **\( n = 4 \)**:
     - For \( n = 4 \), the game involves 8 pieces. A detailed case analysis shows that Andriy can always force Olesya into a losing position by taking pairs that prevent her from accumulating more than half the total weight.
     - Regardless of Olesya's initial move, Andriy can always respond in a way that leaves Olesya with less than half the total weight.

Conclusion:
- Olesya can win if \( n \) is odd or \( n \ge 6 \) and even.
- Olesya cannot win if \( n = 2 \) or \( n = 4 \).

The final answer is \( \boxed{ n \ne 2, 4 } \).