Problem 1506

1. Initial Setup and Definitions:
- We are given a game with initial number $t$ and two positive integers $a$ and $b$ such that $a + b = 2005$.
- Players take turns subtracting either $a$ or $b$ from the current number on the blackboard.
- The player who first reaches a negative number loses the game.

2. Inductive Hypothesis:
- Suppose Player 1 has a winning strategy for some initial number $x$ where $1 \leq x \leq 2005$.
- We need to show that Player 1 also has a winning strategy for $x + 2005k$ for any positive integer $k$.

3. Base Case:
- Assume Player 1 wins for $x$ where $1 \leq x \leq 2005$. Let $P_1$ be the sequence of moves that ensures Player 1's win starting from $x$.

4. Inductive Step:
- Consider the game starting with $t = x + 2005k$.
- Player 1's strategy is to mimic the sequence $P_1$ but adjusted for the larger initial number.
- Specifically, Player 1 starts by subtracting either $a$ or $b$ to reduce the number to $t - a$ or $t - b$.

5. Reduction to Smaller Problem:
- After Player 1's first move, the number on the board is $t - a$ or $t - b$.
- Player 2 then subtracts either $a$ or $b$, resulting in one of the following:
$$t - 2a, \quad t - a - b, \quad t - b - a, \quad t - 2b$$
- Notice that $t - a - b = t - 2005$, which is $x + 2005(k-1)$.

6. Inductive Argument:
- By the inductive hypothesis, if Player 1 has a winning strategy for $x$, then they also have a winning strategy for $x + 2005(k-1)$.
- Therefore, Player 1 can continue to follow the strategy $P_1$ adjusted for the new starting number $x + 2005(k-1)$.

7. Contradiction and Conclusion:
- Suppose $f(t) = 1$ (Player 1 wins starting from $t$).
- If $f(t + a) = 1$ and $f(t + b) = 1$, then $f(t + a + b) = 2$ (Player 2 wins starting from $t + a + b$).
- This implies $f(t + a - b) = 2$ and $f(t + b - a) = 2$, leading to a contradiction since $f(t - a) = 1$ and $f(t - b) = 1$ would imply $f(t) = 2$.

8. Final Proof:
- Therefore, the assumption that $f(t + a + b) \neq f(t)$ leads to a contradiction.
- Hence, $f(t + a + b) = f(t)$ must hold, proving that Player 1 has a winning strategy for infinitely many values of $t$.

$\blacksquare$