62nd Ukrainian National Mathematical Olympiad, Third Round, First Tour

Suppose that more than $674$ numbers were chosen. Then there exists a triple $3n+1,3n+2,3n+3$, among which at least two numbers were chosen, so the absolute difference between some two of the chosen numbers doesn't exceed $2$. Clearly, there exist some two chosen numbers with the same parity, so their sum will be divisible by that difference not exceeding $2$. So, not more than $674$ numbers were chosen.

Let's show that we can choose a given number of numbers. Consider the set $\{1,4,7,\dots ,2020\}$, consisting of $674$ numbers. As the difference between any two of these numbers is divisible by $3$, and the sum of any two of these numbers isn't divisible by $3$, this set satisfies the conditions from the statement.