SAU

What is the greatest number of integers that can be selected from a set of 2015 consecutive numbers so that no sum of any two selected numbers is divisible by their difference?
We take any two chosen numbers. If their difference is 1, it is clear that their sum is divisible by their difference. If their difference is 2, they will be of the same parity, and their sum is divisible by their difference. Therefore, the difference between any chosen numbers will be at least 3. In other words, we can choose at most one number of any three consecutive numbers. This implies that we can choose at most 672 numbers.

Now, we will show that we can choose 672 such numbers from any 2015 consecutive numbers. Suppose that these numbers are $a, a+1, \ldots, a+2014$. If $a$ is divisible by 3, we can choose $a+1, a+4, \ldots, a+2014$. If $a$ is not divisible by 3, we can choose $a, a+3, \ldots, a+2013$.