Lizzie and Alex are playing a game on the whiteboard. Initially, $n$ twos are written
on the board. On a player’s turn they must either 
       1. change any single positive number to 0, or
       2. subtract one from any positive number of positive numbers on the board.
The game ends once all numbers are 0, and the last player who made a move wins. If Lizzie
always plays first, find all $n$ for which Lizzie has a winning strategy.
1. **Initial Claim**: We claim that Lizzie has a winning strategy for all positive integer values of \( n \) such that \( n \not\equiv 0 \pmod{3} \). This means that Lizzie wins if the number of twos on the board is not a multiple of 3.

2. **Base Case**: Consider the base cases:
   - If \( n = 1 \), Lizzie can change the single 2 to 0 and win.
   - If \( n = 2 \), Lizzie can change one of the twos to 0, leaving one 2 for Alex. Alex then changes the remaining 2 to 0, and Lizzie wins.
   - If \( n = 3 \), Lizzie cannot win directly. Any move she makes will leave a winning position for Alex. For example, if Lizzie changes one 2 to 0, Alex is left with two 2s, which is a winning position for Alex.

3. **Inductive Step**: Assume that for all \( k < 3m \) where \( k \not\equiv 0 \pmod{3} \), Lizzie has a winning strategy. We need to show that Lizzie has a winning strategy for \( n = 3m+1 \) and \( n = 3m+2 \).

4. **Case \( n = 3m+1 \)**: 
   - Lizzie can change one of the twos to 0, leaving \( 3m \) twos on the board. By the induction hypothesis, \( 3m \) is a losing position for Alex, so Lizzie wins.

5. **Case \( n = 3m+2 \)**:
   - Lizzie can change one of the twos to 0, leaving \( 3m+1 \) twos on the board. By the induction hypothesis, \( 3m+1 \) is a losing position for Alex, so Lizzie wins.

6. **Conclusion**: By the principle of mathematical induction, Lizzie has a winning strategy for all \( n \not\equiv 0 \pmod{3} \).

The final answer is \( \boxed{ n \not\equiv 0 \pmod{3} } \).