Two players by turns draw diagonals in a regular $(2n+1)$-gon ($n>1$). It is forbidden to draw a diagonal, which was already drawn, or intersects an odd number of already drawn diagonals. The player, who has no legal move, loses. Who has a winning strategy?
[i]K. Sukhov[/i]
1. **Initial Number of Legal Moves**:
   Let's denote the number of legal moves a player can make by \( k \). Initially, the number of diagonals in a regular \((2n+1)\)-gon is given by:
   \[
   k = \frac{(2n+1)(2n)}{2} - (2n+1) = (2n+1)(n-1)
   \]
   This is because the total number of diagonals in a \((2n+1)\)-gon is \(\frac{(2n+1)(2n)}{2} - (2n+1)\), where we subtract the number of sides.

2. **Effect of Drawing a Diagonal**:
   When a player draws a diagonal, it divides the polygon into two parts with \(a\) and \(b\) vertices, where \(a + b = 2n - 1\) (excluding the two endpoints of the diagonal). The number of diagonals between these two parts is \(a \cdot b\), which is an even number since \(a\) and \(b\) are both integers and their product is even.

3. **Intersections with Existing Diagonals**:
   When a diagonal is drawn, it intersects an even number of already drawn diagonals. This is because the total number of intersections (including those with not yet drawn diagonals) is even, and the number of intersections with not yet drawn diagonals is also even.

4. **Parity Change in Legal Moves**:
   Let \(u_1\) be the number of diagonals that have an odd number of intersections with already drawn diagonals, and \(u_2\) be the number of diagonals that have an even number of intersections. Initially, \(u_1 + u_2\) is even. After drawing a diagonal, \(u_1\) will become even and \(u_2\) will become odd. The number of legal moves \(k\) changes as follows:
   \[
   k \rightarrow k + u_1 - u_2 - 1
   \]
   Since \(u_1 - u_2\) is even, the parity of \(k\) changes.

5. **Winning Strategy**:
   The player who has an odd number of legal moves will always have an odd number of options, while the other player will have an even number of options. This means the player with an even number of options will eventually have no legal move and lose.

6. **Initial Parity of \(k\)**:
   - If \(n\) is even, then the initial \(k\) is odd, so the first player will win.
   - If \(n\) is odd, then the initial \(k\) is even, so the second player will win.

\(\blacksquare\)

The final answer is: If \( \boxed{ n } \) is even, the first player wins; if \(n\) is odd, the second player wins.