BMO Short List

We claim that the maximum possible value is $3029$. The polynomials $$P(x)=(x^2-1{)}^{1009}(x^2+1)\text{and}Q(x)=(x^2-1{)}^{1009}(x^2+x+1)$$ satisfy the conditions, have $2018$ common roots, and have $1011$ common coefficients (all coefficients of even powers). So $r+s\ge 3029$.

Suppose now that $P(x)$, $Q(x)$ agree on the coefficients of $x^{2020}$, $x^{2019}$, $\dots$, $x^{2021-k}$, disagree on the coefficient of $x^{2020-k}$, and agree on another $s-k$ coefficients. The common roots of $P(x)$, $Q(x)$ are also non-zero roots of the polynomial $P(x)-Q(x)$ which has degree $x^{2020-k}$. (The condition on the non-zero coefficients guarantees that $0$ is not a root of $P$ and $Q$.) So $P(x)-Q(x)$ has at most $2020-k$ real roots. On the other hand, $P(x)-Q(x)$ has exactly $s-k$ coefficients equal to zero. So by the following Lemma it has at most $2[(2020-k)-(s-k)]=4040-2s$ real non-zero roots.

Averaging we get $r\le \lfloor \frac{6060-2s-k}{2}\rfloor \le 3030-s$. Thus $r+s\le 3030$. Furthermore, if equality occurs, we must have $k=0$ and $r=2020-k=4040-2s$. In other words, we must have $r=2020$ and $s=1010$. But if $r=2020$, then $Q(x)$ is a multiple of $P(x)$ and since $P(x)$, $Q(x)$ have non-zero coefficients, then $s=0$, a contradiction. Therefore $r+s\le 3029$ as required.