Problem 1418

Let's analyze the game for each value of $n$ given in the problem.

### Case a) $n = 2019$

1. **Determine the parity of $n$**:
$$n = 2019 \quad \text{(odd)}$$
Since $n$ is odd, the first player $A$ will make the last move.

2. **Strategy for player $A$**:
- Player $A$ starts by writing a 1.
- Player $B$ must write a 0 to prevent $A$ from winning immediately.
- Player $A$ continues by writing a 0.
- Player $B$ must write a 1 to prevent $A$ from winning immediately.

This pattern continues, and the sequence of numbers will be:
$$1, 0, 0, 1, 1, 0, 0, 1, \ldots, 1, 0, 0, 1, 1$$

3. **Winning condition for player $A$**:
- After $A$'s last move, the sequence will have a pattern where there are three consecutive vertices with the same number (either 1 or 0).
- For example, the first, second, and $(2k+1)$-th vertices will form a triangle with the sum divisible by 3.

Thus, player $A$ has a winning strategy when $n = 2019$.

### Case b) $n = 2020$

1. **Determine the parity of $n$**:
$$n = 2020 \quad \text{(even)}$$
Since $n$ is even, the second player $B$ will make the last move.

2. **Strategy for player $B$**:
- Player $A$ starts by writing a 1.
- Player $B$ writes a 0 to prevent $A$ from winning immediately.
- Player $A$ continues by writing a 0.
- Player $B$ writes a 1 to prevent $A$ from winning immediately.

This pattern continues, and the sequence of numbers will be:
$$1, 0, 0, 1, 1, 0, 0, 1, \ldots, 1, 0, 0, 1, 1$$

3. **Winning condition for player $B$**:
- Player $B$ ensures that there are no three consecutive vertices with the same number.
- Since $B$ makes the last move, $A$ cannot form a triangle with the sum divisible by 3.

Thus, player $B$ has a winning strategy when $n = 2020$.

### Case c) $n = 2021$

1. **Determine the parity of $n$**:
$$n = 2021 \quad \text{(odd)}$$
Since $n$ is odd, the first player $A$ will make the last move.

2. **Strategy for player $A$**:
- Player $A$ starts by writing a 1.
- Player $B$ must write a 0 to prevent $A$ from winning immediately.
- Player $A$ continues by writing a 0.
- Player $B$ must write a 1 to prevent $A$ from winning immediately.

This pattern continues, and the sequence of numbers will be:
$$1, 0, 0, 1, 1, 0, 0, 1, \ldots, 1, 0, 0, 1, 1$$

3. **Winning condition for player $A$**:
- After $A$'s last move, the sequence will have a pattern where there are three consecutive vertices with the same number (either 1 or 0).
- For example, the first, second, and $(2k+1)$-th vertices will form a triangle with the sum divisible by 3.

Thus, player $A$ has a winning strategy when $n = 2021$.

The final answer is: