OFM 2021 S2

Albert and Beatrice play a game. $2021$ stones lie on a table. Starting with Albert, they alternatively remove stones from the table, while obeying the following rule. At the $n$ -th turn, the active player (Albert if $n$ is odd, Beatrice if $n$ is even) can remove from $1$ to $n$ stones. Thus, Albert first removes $1$ stone; then, Beatrice can remove $1$ or $2$ stones, as she wishes; then, Albert can remove from $1$ to $3$ stones, and so on. The player who removes the last stone on the table loses, and the other one wins. Which player has a strategy to win regardless of the other player's moves?