Two players in turns color the sides of an $n$-gon. The first player colors any side that has $0$ or $2$ common vertices with already colored sides. The second player colors any side that has exactly $1$ common vertex with already colored sides. The player who cannot move, loses. For which $n$ the second player has a winning strategy?
1. **Base Case Analysis:**
   - For \( n = 4 \):
     - The first player colors any side, say side 1.
     - The second player can then color any side that shares exactly one vertex with side 1, say side 2.
     - The first player now cannot color any side because all remaining sides share exactly one vertex with already colored sides.
     - Therefore, the first player loses, and the second player wins.
   - For \( n = 3 \):
     - The first player colors any side, say side 1.
     - The second player cannot color any side because all remaining sides share exactly one vertex with side 1.
     - Therefore, the second player loses, and the first player wins.
   - For \( n = 5 \):
     - The first player colors any side, say side 1.
     - The second player can color any side that shares exactly one vertex with side 1, say side 2.
     - The first player can then color a side that shares no vertices with side 2, say side 3.
     - The second player can color a side that shares exactly one vertex with side 3, say side 4.
     - The first player can then color the remaining side, say side 5.
     - Therefore, the first player wins.

2. **Inductive Step:**
   - Assume the second player has a winning strategy for \( n = k \).
   - We need to show that the second player has a winning strategy for \( n = k + 3 \).
   - Label the edges from 1 to \( k + 3 \) cyclically.
   - The first player colors edge \( k + 3 \).
   - The second player colors edge \( k + 2 \), which shares exactly one vertex with edge \( k + 3 \).
   - The first player colors edge \( k \), which shares no vertices with edges \( k + 2 \) and \( k + 3 \).
   - The second player now erases edges \( k \), \( k + 2 \), and \( k + 3 \) and replaces them with a single edge \( k \).
   - The game is now reduced to \( k \) edges, and by the induction hypothesis, the second player has a winning strategy for \( k \).

3. **Conclusion:**
   - By induction, the second player has a winning strategy for \( n = 4 \) and for all \( n \) of the form \( n = 4 + 3k \) for \( k \geq 0 \).

The final answer is \( \boxed{ n = 4 + 3k } \) for \( k \geq 0 \).