Denote: Albert=player $1$;Beatrice=player $0$. Consider two consecutive turns $k$ and $k+1$, in which the players remove $s_k$, respectively $s_{k+1}$ stones. With the notation $m=k+1\pmod{2}$, the player $m$ can always obtain each of the possibilities $s_k+s_{k+1}=k+1$ or $s_k+s_{k+1}=k+2$. Proof: Let be $s_k=p\in[1,k]$. Then $k+1-p\in[1,k]\subset[1,k+1];k+2-p\in[2,k+1]\subset[1,k+1]$, hence exists $s'_{k+1}\in[1,k+1]$ such that $s_k+s'_{k+1}=k+1$ and exists $s''_{k+1}\in[1,k+1]$ such that $s_k+s''_{k+1}=k+2$. This conclusion is favorable to the player $0$ (Beatrice). Denote $S_k=\sum_{i=1}^ks_i$. The winner is the player which can remove the $2020$-th stone, hence the player which can obtain $S_k=2020$. For $n\ge1$, Beatrice can always obtain each of the possibilities $s_{2n-1}+s_{2n}=2n$ or $s_{2n-1}+s_{2n}=2n+1$. Hence, Beatrice can obtain $\sum_{n=1}^N2n\le S_{2N}\le\sum_{n=1}^N(2n+1)\Longrightarrow (N+1)^2-(N+1)\le S_{2N}\le (N+1)^2-1$. For $N=44$ results: $1980\le S_{88}\le 2024$ and $2020\in[1980,2024]$. Hence, Beatrice can play such that $S_{88}=2020$: she obtains $4$ times $s_{2n-1}+s_{2n}=2n$ and $40$ times $s_{2n-1}+s_{2n}=2n+1$.