HMMT February

Let $a=2021$ and $b=2000$. It is clear that we may increase $y-x$ unless both $x-1$ and $y+1$ are multiples of $b$, so we may assume that our interval is of length $b(k+1)-1$, where there are $k$ multiples of $b$ in our interval. There are at least $k+1$ multiples of $a$, and so it is of length at least $a k+1$. We thus have that $$a k+1 \leq b(k+1)-1 \Longrightarrow(a-b) k \leq b-2 \Longrightarrow k \leq\left\lfloor\frac{b-2}{a-b}\right\rfloor$$ So, the highest possible value of $k$ is 95, and this is achievable by the Chinese remainder theorem, giving us an answer of 191999.