To determine for which values of $N$ player $B$ wins, we need to analyze the structure of the game and identify a strategy that ensures victory for player $B$.

### Game Analysis

Given the rules of the game:
- Player $A$ starts by writing the number $1$.
- Each player alternates turns writing either $n+1$ or $2n$, where $n$ is the last number written.
- The number written cannot exceed $N$.
- The player who writes $N$ wins.

To solve the problem, we must deduce for which values of $N$, player $B$ is guaranteed a win regardless of how optimally player A \ plays. ### Winning and Losing Positions In combinatorial game theory, we identify "winning" and "losing" positions: - **Winning Position:** A position where the player whose turn it is can force a win with optimal play. - **Losing Position:** A position where the player whose turn it is will lose with optimal play from the opponent. The strategy involves determining the losing positions. Player \( B will win if and only if $A$ begins his turn in a losing position.

### Characterizing Losing Positions

1. Base Case:
- $N = 1$ is a losing position because player $A$ starts and immediately wins.

2. Recursive Analysis:
- If writing $n$ is a losing position, then both $n+1$ and $2n$ must be winning positions.
- Conversely, $n$ is a winning position if either $n+1$ or $2n$ is a losing position.

By iteratively applying these conditions, we can deduce:
- A position $n$ is losing if $n$ is the sum of distinct odd powers of 2. These are numbers whose binary representation consists of zeros and a single one in an odd position.

### Examples and Verification

For example, consider $N = 3$.
- $N = 1$ is losing (as defined above).
- Starting from 1, player $A$ can choose 2.
- Then, player $B$ can choose 3 and win. Hence, 3 was a winning position for $B$.

Values of $N$ being sums of distinct odd powers of 2 include $N = 1, 3, 5, 9, 17, \ldots$. In binary, these are 1, 11, 101, 1001, 10001, corresponding to $2^0, 2^1 + 2^0, 2^2 + 2^0, 2^3 + 2^0, 2^4 + 2^0, \ldots$.

### Final Conclusion

Player $B$ wins for values of $N$ that are sums of distinct odd powers of 2. Therefore, the answer is:

$$\boxed{N = \text{the sum of distinct odd powers of } 2}$$

This strategy ensures that player $B$ will always find themselves in a winning position when $N$ fits this pattern. The correct identification of such $N$ solidifies player $B$'s chances of winning.