, 2015

Let $n$ be an arbitrary integer and
$$S_n = \{n - 2015, n - 2014, \dots, n, \dots, n + 2015\}$$
be the set of all integers that differ from $n$ by at most $2015$.
We know that $S_{f(n)} = \{f(n)-2015, f(n)-2014, \dots, f(n)+2015\}$.
From the condition of the problem it follows that the numbers $f(n-2015), f(n-2014), \dots, f(n+2015)$ are also elements of the set $S_{f(n)}$, since they differ from $f(n)$ by at most $2015$. Because $f$ is injective all these numbers are different. There are $4031$ of them, which is also the size of the set $S_{f(n)}$. Thus
$$S_{f(n)} = \{f(n) - 2015, \dots, f(n) + 2015\} = \{f(n - 2015), \dots, f(n + 2015)\}$$

Let us now look at the intersection of the sets $S_{f(n)}$ and $S_{f(n+1)}$. On one hand we have
$$\begin{aligned} S_{f(n)} \cap S_{f(n+1)} &= \{f(n - 2015), \dots, f(n + 2015)\} \cap \{f(n - 2014), \dots, f(n + 2016)\} \\ &= \{f(n - 2014), \dots, f(n + 2015)\}. \end{aligned}$$
So the intersection has $4030$ elements. On the other hand we have
$$S_{f(n)} \cap S_{f(n+1)} = \{f(n) - 2015, \dots, f(n) + 2015\} \cap \{f(n+1) - 2015, \dots, f(n+1) + 2015\}.$$
This set can have $4030$ elements if and only if $f(n)$ and $f(n+1)$ differ by $1$. Thus we proved that $|f(n+1) - f(n)| = 1$ for every integer $n$.
If for some $n$ we have $f(n+1) - f(n) = 1$ then by induction we have $f(n+k) - f(n) = k$ for every integer $k$. Similarly, if for some $n$ we have $f(n+1) - f(n) = -1$ then we have $f(n+k) - f(n) = -k$ for every integer $k$. In both cases we get $|f(m) - f(n)| = |m-n|$ for all integers $m$ and $n$.