From this post, we conclude there are no solutions. Also, this is similar to IMO 1987 P4.

I'm going to write the proof refer to IMO 1987 P4. Suppose that there is such f.Let A=N－f（N）, B=f（A）.Apparently f is injective.We claim that B=f（N）－f（f（N））.Because if k∈B, k∈f（N） and from injectivity of f, k∈f（N）－f（f（N））.On the other hand, if k∈f（N）－f（f（N））, k∈f（A）=B. Since A and B are disjoint, |A∪B|=|A|＋|B|.And from injectivity of f, |A|=|B|, so |A∪B| is even.On the other hand, A∪B=N－f（f（N））=｛1,...,2015｝ which is absurd. Therefore there is no such solution f.

It can be easily seen that the function $f$ is injective Our problem function produces natural values from $2016$ onwards Suppose the function $f$ alone does not generate $N$ values. So the function $f(f(n))$ does not produce $2N$ value. So, if a function wants to be true in the problem, the following equation must have a solution $2N=2015$ And this issue has no natural answer! So the function does not work in the problem