Answer. $\boxed{m=2023r.}$ That works since there exists set containing $a=2022r+x$ and $b=2022r+y$ for $0\leq x<y\leq r$. Indeed, $$\frac{2022r+y}{2022r+x}\leq 1+\frac{x-y}{2022r+x}\leq 1+\frac{x-y}{2022r}\leq 1+\frac{r}{2022r}=1+\frac{1}{2022}.$$ Consider $m<2023r$ and suppose that statement is still true. Consider sets $A_i=\{i\pmod{r}\}$. For any such set, $b-a\geq r$. Hence, $$\frac{1}{2022}\geq\frac{b-a}{a}\geq \frac{r}{a}\implies m\geq b\geq a+r\geq 2023r,$$which yields contradiction. $\blacksquare$