Assume that we have chosen $a$ and $b$ where $a>b$. If $a-b=1$, then $a-b|a+b$. contradiction If $a-b=2$, then $2|2b+2=a+b$. contradiction Thus $a-b\ge 3$. Let $x_1,x_2,...,x_n$ where $n\in Z_+$ be the chosen numbers in increasing order. Then $x_n\ge x_{n-1}+3\ge...\ge x_1+3(n-1)$. Because this string is contained in subset of consecutive $2015$ integers, then $2015-1\ge x_n-x_1\ge 3(n-1)\implies 672\ge n$. Now we prove there exist such sequence for $n=672$ which will show that sought greatest number of integers is $672$. For example set $\lbrace 1,2,...,2015\rbrace$ and sequence $x_k=3k-1$ for all $k\in\lbrace 1,2,..., 672$. Then difference of every two is divisible by $3$ but their sum is not.