3. Let $n$ be a positive integer no less than 3, and write it on the blackboard as the initial value. Two players, A and B, take turns to modify the number on the blackboard. A can add 3 or subtract 1 each time, while B can add 2 or subtract 2 each time. When someone writes $k$, such that $|k-n| \geqslant n$, the player wins and the game ends. If a number written in the current round has appeared before in the same round, it is a draw. If a draw occurs, the next round starts with the initial value being 1 less than the initial value of the previous round. B decides who goes first in the first round, and in subsequent rounds (if any), the player who was supposed to go next but did not in the previous round goes first. Question: Who has a winning strategy?
3. A must win.

When $n=3$, it is easy to see that, regardless of who modifies first, A will win on their first turn.

When $n=4$, if A goes first, the possible game progressions are $4 \rightarrow 3 \rightarrow 5 \rightarrow 8$ or $4 \rightarrow 3 \rightarrow 1 \rightarrow 0$, A wins; if B goes first, the possible game progressions are $4 \rightarrow 6 \rightarrow 9$, A wins, $4 \rightarrow 2 \rightarrow 5 \rightarrow 7 \rightarrow 10$, A wins, or $4 \rightarrow 2 \rightarrow 5 \rightarrow 3 \rightarrow 2$, a draw, and the next game starts with 3, then A must win.

When $n \geqslant 5$, if B goes first, it is clear that B cannot win directly, so A will have a turn.
Let the number on the board when A has a turn be $k$.
If $k \geqslant 2 n-3$, then A wins;
If $k \leqslant n-2$, then A changes it to $k+3$, by
$$
4 \leqslant k+3 \leqslant n+1 \text {, }
$$

so B cannot win directly after their turn;
If $n-1 \leqslant k \leqslant 2 n-4$, then A changes it to $k-1$, by
$$
n-2 \leqslant k-1 \leqslant 2 n-5 \text {, }
$$

so B cannot win directly after their turn either.
In summary, B has no winning strategy. If a draw occurs, the initial value of the next game will decrease. By this reasoning, when $n$ decreases to 4, A will win.