Solution:
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 $a, a+3, \ldots, a+2013$.