Solution:

We first prove the following lemma.

LEMMA. If in the $(t ; a, b)$-game some of the two players has a winning strategy, then in the $(t+a+b, a, b)$-game the same player has a winning strategy.

Proof of the lemma. Denote the players by $A$ and $B$ and let $B$ have a winning strategy for the $(t ; a, b)$-game. In the $(t+a+b ; a, b)$-game after the first move of $A$ we obtain either the $(t+a ; a, b)$-game or the $(t+b ; a, b)$-game with $B$ to go first. In both cases $B$ can get the $(t ; a, b)$-game with $A$ as first player in which case $B$ has a winning strategy.

Let us now assume that $A$ has a winning strategy for the $(t ; a, b)$-game. Then after the first move of $A$ we obtain either the $(t-a ; a, b)$-game or the $(t-b ; a, b)$-game with $B$ to go first. Therefore the second player has a winning strategy for some of these games.

We consider (without loss of generality) the case of the $(t-a ; a, b)$-game. It follows from the above that the second player has a winning strategy for the $(t-a+a+b=t+b ; a, b)$-game. Since $A$ can obtain the $(t+b ; a, b)$-game with $B$ to go first from the $(t+a+b ; a, b)$-game, it follows that $A$ has a winning strategy for the $(t+a+b ; a, b)$-game. This completes the proof of the lemma.

We now prove that for $t=2004$ and $a+b=2005$ the first player $A$ has a winning strategy. We may assume that $a \leq b$. Since $a>0$, we have $b \leq 2004$. Then $A$ subtracts $b$ from $t=2004$. The resulting number $2004-b$ is less than $a$ since $a+b=2005$. This means that any move of $B$ leads to a negative number. Now the lemma implies that $A$ has a winning strategy for the $(t, a, b)$-game for every $t \equiv 2004 \pmod{2005}$ and $a+b=2005$.