Let $k \geq 1$ and $N>1$ be two integers. On a circle are placed $2N+1$ coins all showing heads. Calvin and Hobbes play the following game. Calvin starts and on his move can turn any coin from heads to tails. Hobbes on his move can turn at most one coin that is next to the coin that Calvin turned just now from tails to heads. Calvin wins if at any moment there are $k$ coins showing tails after Hobbes has made his move. Determine all values of $k$ for which Calvin wins the game. 

[i]Proposed by Tejaswi Navilarekallu[/i]
1. **Initial Setup and Definitions**:
   - Let \( k \geq 1 \) and \( N > 1 \) be two integers.
   - There are \( 2N + 1 \) coins on a circle, all initially showing heads.
   - Calvin and Hobbes play a game where Calvin can turn any coin from heads to tails, and Hobbes can turn at most one adjacent coin from tails to heads.
   - Calvin wins if at any moment there are \( k \) coins showing tails after Hobbes has made his move.

2. **Strategy for Hobbes when \( k \geq N + 2 \)**:
   - Hobbes pairs the coins \( (2i-1, 2i) \) for \( 1 \leq i \leq N \).
   - If Calvin turns both coins in a pair to tails, Hobbes turns one of them back to heads.
   - This ensures that no pair has more than one tail.
   - Therefore, the number of tails after Hobbes' move is at most \( 1 + \frac{(2N+1) - 1}{2} = N + 1 \).
   - Hence, Hobbes wins if \( k \geq N + 2 \).

3. **Strategy for Calvin when \( k \leq N \)**:
   - Calvin can simply turn coins \( 2i \) for \( 1 \leq i \leq N \) to tails.
   - This results in \( N \) tails, and since \( k \leq N \), Calvin wins.

4. **Case Analysis for \( k = N + 1 \)**:
   - Let \( N = 2m + \varepsilon \) where \( \varepsilon \in \{0, 1\} \).
   - Consider \( m \) arcs on the circle, each containing 4 coins: \( \{4i+1, 4i+2, 4i+3, 4i+4\} \) for \( 0 \leq i < m \).

5. **Calvin's Moves**:
   - Calvin makes \( 3m \) moves:
     - On move \( 3i+1 \), he turns coin \( 4i+2 \) to tails.
     - On move \( 3i+2 \), he turns coin \( 4i+4 \) to tails.
     - On move \( 3i+3 \), he turns coin \( 4i+3 \) to tails.
   - After these moves, each arc will have either:
     - Type \( \widehat{23} \): \( \{4i+2, 4i+3\} \) tails.
     - Type \( \widehat{34} \): \( \{4i+3, 4i+4\} \) tails.
     - Type \( \widehat{234} \): \( \{4i+2, 4i+3, 4i+4\} \) tails.

6. **Case 1: \( \varepsilon = 0 \)**:
   - If any arc is of type \( \widehat{234} \), there are \( \geq 3 + 2(m-1) = N + 1 \) tails, and Calvin wins.
   - Assume all arcs are of type \( \widehat{23} \) or \( \widehat{34} \), resulting in \( 2m \) tails.
   - If coin \( 4m+1 \) has no tails neighbors, Calvin turns it to win.
   - If the arc \( \{4m-3, 4m-2, 4m-1, 4m\} \) is of type \( \widehat{34} \):
     - If \( \{1, 2, 3, 4\} \) is also of type \( \widehat{34} \), Calvin turns coin \( 1 \) to win.
     - If \( \{1, 2, 3, 4\} \) is of type \( \widehat{23} \), there exists an \( i \) such that the \( i \)-th arc is of type \( \widehat{23} \) and the \( (i+1) \)-th arc is of type \( \widehat{34} \). Calvin turns coin \( 4i+1 \) to win.

7. **Case 2: \( \varepsilon = 1 \)**:
   - If any arc is of type \( \widehat{234} \), Calvin turns coin \( 4m+2 \) and wins with \( \geq 3 + 2(m-1) + 1 = N + 1 \) tails.
   - Assume all arcs are of type \( \widehat{23} \) or \( \widehat{34} \), resulting in \( 2m \) tails.
   - If Calvin can turn coin \( 4m+1 \), he wins by turning \( 4m+3 \) next move.
   - If the arc \( \{4m-3, 4m-2, 4m-1, 4m\} \) is of type \( \widehat{34} \):
     - If \( \{1, 2, 3, 4\} \) is of type \( \widehat{34} \), Calvin turns coin \( 1 \) and \( 4m+2 \) to win.
     - If \( \{1, 2, 3, 4\} \) is of type \( \widehat{23} \), there exists an \( i \) such that the \( i \)-th arc is of type \( \widehat{23} \) and the \( (i+1) \)-th arc is of type \( \widehat{34} \). Calvin wins by turning \( 4m+2 \) and \( 4i+1 \) in the next two moves.

\(\blacksquare\)

The final answer is \( \boxed{ k \in \{1, 2, \dots, N+1\} } \).