The answer is : they can force a tie. Firstofall, let $S_A(k)$ and $S_B(k)$ the number of points corresponding to Ana and Benito in the $k-$th turn. We will prove the following claim : Claim.In his turn, Benito can increase $S_B(k)-S_A(k)$ by at most $1$. Proof.It is pretty clear that he can add as many consecutive numbers as he wants to $S_A(k)$, but can only add one to $S_B(k)$, so $S_B(k)-S_A(k)$ can be increased by $k+l-(k+(k+1)+\dots+(k+l-1)=(1-l)k+(l-\frac{l(l-1)}{2}) \le 1$ (because $l \ge 1$, since the strategy of adding $k$ to his sum won't work, the reason being the fact that Ana could choose to add $k+1$ to her sum, and this would decrease $S_B(k)-S_A(k)$ by $1$). The equality is attained when Benito adds $k$ to Ana's sum and $k+1$ to his. $\square$ Similarly, Ana can increase $S_A(k)-S_B(k)$ by at most $1$, and since $2020$ is a multiple of $4$, we get that $S_A(2020)=S_B(2020)$, which proves that both of them can force a tie. $\blacksquare$ Remark. I think the part of the question about a tie shouldn't be there, because it suggests the answer