Solved with L567. We essentially want the smallest $b$ such that there exists an $a < b$ with $a(a+1) \equiv b(b+1) \pmod {2021}$.Notice how $(a, b) \equiv (19, 66)$ works. Now to win, notice that $a^2+a \equiv b^2+b \pmod {2021}$ if and only if $2021 \mid (a-b)(a+b+1)$. Since $2021 = 43 \cdot 47$, both of which are prime it is very easy to do a little bashing and assert that $66$ is indeed the minimal such $b$. Thus the answer should be $66$.