Alex the Kat and Kelvin the Frog play a game on a complete graph with $n$ vertices. Kelvin goes first, and the players take turns selecting either a single edge to remove from the graph, or a single vertex to remove from the graph. Removing a vertex also removes all edges incident to that vertex. The player who removes the final vertex wins the game. Assuming both players play perfectly, for which positive integers $n$ does Kelvin have a winning strategy?
1. **Base Case:**
   - Consider the base case where \( n = 0 \). In this case, there are no vertices to remove, so Kelvin loses by default. This establishes that Kelvin loses for \( K_0 \).

2. **Inductive Step:**
   - Assume that Kelvin loses for \( K_n \). We need to show that:
     - Kelvin wins for \( K_{n+1} \).
     - Kelvin wins for \( K_{n+2} \).
     - Kelvin loses for \( K_{n+3} \).

3. **Case \( K_{n+1} \):**
   - Kelvin can remove a vertex from \( K_{n+1} \), reducing it to \( K_n \). Since we assumed Kelvin loses for \( K_n \), Alex has a winning strategy for \( K_n \). Kelvin can now adopt Alex's strategy for \( K_n \) and win \( K_{n+1} \).

4. **Case \( K_{n+2} \):**
   - Kelvin can remove an edge, say \( ab \), from \( K_{n+2} \). This leaves a graph that is still essentially \( K_{n+2} \) but with one less edge. Kelvin can now use Alex's strategy for \( K_n \). If Alex's strategy involves removing a vertex or edge that affects \( a \) or \( b \), Kelvin can mirror the move on the other vertex, maintaining symmetry and ensuring he wins.

5. **Case \( K_{n+3} \):**
   - If Kelvin removes a vertex, the game reduces to \( K_{n+2} \). Since we assumed Kelvin wins for \( K_{n+2} \), Alex can now adopt Kelvin's strategy for \( K_{n+2} \) and win.
   - If Kelvin removes an edge, say \( ab \), the game reduces to \( K_{n+1} \). Since we assumed Kelvin wins for \( K_{n+1} \), Alex can now adopt Kelvin's strategy for \( K_{n+1} \) and win by maintaining symmetry with respect to \( a \) and \( b \).

6. **Conclusion:**
   - By induction, if Kelvin loses for \( K_n \), he wins for \( K_{n+1} \) and \( K_{n+2} \), and loses for \( K_{n+3} \).
   - Since Kelvin loses for \( K_0 \), it follows by induction that Kelvin wins for all \( n \) that are not multiples of 3, and loses for all \( n \) that are multiples of 3.

The final answer is \( \boxed{ n } \) is not divisible by 3.