Equivalently, $(m,n)$ is special if $m-n \mid 60m$. In particular, if $\vert m-n\vert \in \{1,2,3,4,5,6\}$. So any two numbers we choose need to have difference at least $7$. In particular, we can choose at most $\left\lceil \frac{2021}{7}\right\rceil=289$ numbers. On the other hand, just choosing the numbers $1+7k$ for $k=0,1,2\dots,288$ the difference of any two numbers is a multiple of $7$ and hence does not divide $60m$ since $m$ is coprime to $7$. Hence the answer: $289$.