In a heap there are $2021$ stones. Two players $A$ and $B$ play removing stones of the pile, alternately starting with $A$. A valid move for $A$ consists of remove $1, 2$ or $7$ stones. A valid move for B is to remove $1, 3, 4$ or $6$ stones. The player who leaves the pile empty after making a valid move wins. Determine if some of the players have a winning strategy. If such a strategy exists, explain it.
1. **Identify the losing positions for player A:**
   - We need to determine the positions where player A cannot make a move that forces player B into a losing position.
   - Let's analyze the positions from 1 to 10:
     - If there is 1 stone, A can remove 1 stone and win.
     - If there are 2 stones, A can remove 2 stones and win.
     - If there are 3 stones, A can remove 2 stones, leaving 1 stone for B, who can only remove 1 stone and lose.
     - If there are 4 stones, A can remove 2 stones, leaving 2 stones for B, who can only remove 1, 3, or 4 stones. If B removes 1 stone, A can remove the remaining 1 stone and win. If B removes 3 stones, A can remove the remaining 1 stone and win.
     - If there are 5 stones, A can remove 1, 2, or 7 stones. Removing 1 stone leaves 4 stones for B, who can remove 1, 3, 4, or 6 stones. Removing 2 stones leaves 3 stones for B, who can remove 1, 3, 4, or 6 stones. Removing 7 stones is not possible. In all cases, B can make a move that leaves A in a winning position.
     - If there are 6 stones, A can remove 1, 2, or 7 stones. Removing 1 stone leaves 5 stones for B, who can remove 1, 3, 4, or 6 stones. Removing 2 stones leaves 4 stones for B, who can remove 1, 3, 4, or 6 stones. Removing 7 stones is not possible. In all cases, B can make a move that leaves A in a winning position.
     - If there are 7 stones, A can remove 1, 2, or 7 stones. Removing 1 stone leaves 6 stones for B, who can remove 1, 3, 4, or 6 stones. Removing 2 stones leaves 5 stones for B, who can remove 1, 3, 4, or 6 stones. Removing 7 stones leaves 0 stones, and A wins.
     - If there are 8 stones, A can remove 1, 2, or 7 stones. Removing 1 stone leaves 7 stones for B, who can remove 1, 3, 4, or 6 stones. Removing 2 stones leaves 6 stones for B, who can remove 1, 3, 4, or 6 stones. Removing 7 stones leaves 1 stone, and A wins.
     - If there are 9 stones, A can remove 1, 2, or 7 stones. Removing 1 stone leaves 8 stones for B, who can remove 1, 3, 4, or 6 stones. Removing 2 stones leaves 7 stones for B, who can remove 1, 3, 4, or 6 stones. Removing 7 stones leaves 2 stones, and A wins.
     - If there are 10 stones, A can remove 1, 2, or 7 stones. Removing 1 stone leaves 9 stones for B, who can remove 1, 3, 4, or 6 stones. Removing 2 stones leaves 8 stones for B, who can remove 1, 3, 4, or 6 stones. Removing 7 stones leaves 3 stones, and A wins.

   From this analysis, we see that the losing positions for A are 5 and 10.

2. **Determine the strategy for player A:**
   - Player A can force player B into a losing position by making moves that leave the number of stones in a position that is a multiple of 5.
   - If A can always leave a number of stones that is 0 or 2 modulo 5, B will be forced into a losing position.

3. **Apply the strategy to the initial number of stones:**
   - The initial number of stones is 2021.
   - We need to check if 2021 modulo 5 is 0 or 2:
     \[
     2021 \mod 5 = 1
     \]
   - Since 2021 modulo 5 is 1, A can remove 1 stone, leaving 2020 stones.
   - Now, 2020 modulo 5 is 0, which is a losing position for B.

4. **Conclusion:**
   - Player A can always force player B into a losing position by making the right moves.

The final answer is $\boxed{A\; wins}$