There are $N$ piles each consisting of a single nut. Two players in turns play the following game. At each move, a player combines two piles that contain coprime numbers of nuts into a new pile. A player who can not make a move, loses. For every $N > 2$ determine which of the players, the first or the second, has a winning strategy.
1. **Initial Configuration and Lemma**:
   - We start with \( N \) piles, each containing 1 nut.
   - We need to prove that for \( N > 2 \), the second player has a winning strategy.
   - Consider the lemma: If a player is to move on a configuration that contains \( k \ge 2 \) piles of 1 nut each and one pile of an odd number \( t \) of nuts, denoted as \([k \times 1, t]\), he cannot prevent that he will be confronted on his next move with the configuration \([(k-2) \times 1, (t+2)]\) if his opponent so desires.

2. **Proof of Lemma**:
   - The player can either:
     - Combine two 1-nut piles, resulting in \([(k-2) \times 1, 2, t]\).
     - Combine the \( t \)-nut pile with a 1-nut pile, resulting in \([(k-1) \times 1, (t+1)]\).
   - In the first case, the opponent can combine the 2-nut pile with the \( t \)-nut pile, resulting in \([(k-2) \times 1, (t+2)]\).
   - In the second case, the opponent can combine a 1-nut pile with the \((t+1)\)-nut pile, resulting in \([(k-2) \times 1, (t+2)]\).
   - Since \( t \) is odd, 2 and \( t \) are coprime, and there is always a 1-nut pile available for the second case.

3. **Application of Lemma for Odd \( N \)**:
   - If \( N \) is odd, the starting configuration \([N \times 1]\) can be written as \([(N-1) \times 1, 1]\).
   - By the lemma, player 2 can ensure that player 1 faces configurations \([(N-3) \times 1, 3]\), \([(N-5) \times 1, 5]\), ..., \([2 \times 1, (N-2)]\), \([N]\).
   - On the last configuration \([N]\), player 1 cannot make a legal move, and thus loses.

4. **Application of Lemma for Even \( N \)**:
   - If \( N \) is even, the starting configuration \([N \times 1]\) can be reduced to \([3 \times 1, (N-3)]\) by player 2.
   - At \([3 \times 1, (N-3)]\), player 1's options are:
     - Combine two 1-nut piles, resulting in \([1, 2, (N-3)]\).
     - Combine a 1-nut pile with the \((N-3)\)-nut pile, resulting in \([2 \times 1, (N-2)]\).
   - In the first case, player 2 combines the 2-nut pile with the \((N-3)\)-nut pile, resulting in \([2, (N-2)]\).
   - In the second case, player 2 combines a 1-nut pile with the \((N-2)\)-nut pile, resulting in \([2, (N-2)]\).
   - Since \( N \) is even, player 1 cannot make a legal move from \([2, (N-2)]\), and thus loses.

Conclusion:
- For \( N > 2 \), the second player has a winning strategy by ensuring that the first player is always confronted with configurations that lead to a losing position.

The final answer is the second player has a winning strategy for \( \boxed{ N > 2 } \).