jmc

Let $a=f(1)$ and $b=f(2).$ Then $$\begin{array}{rl}f(3)& =f(2)-f(1)=b-a,\\ f(4)& =f(3)-f(2)=(b-a)-b=-a,\\ f(5)& =f(4)-f(3)=-a-(b-a)=-b,\\ f(6)& =f(5)-f(4)=-b-(-a)=a-b,\\ f(7)& =f(6)-f(5)=(a-b)-(-b)=a,\\ f(8)& =f(7)-f(6)=a-(a-b)=b.\end{array}$$Since $f(7)=f(1)$ and $f(8)=f(2),$ and each term depends only on the previous two terms, the function becomes periodic from here on, with a period of length 6.

Then $f(3)=f(15)=20$ and $f(2)=f(20)=15,$ and $$f(20152015)=f(1)=f(2)-f(3)=15-20=\boxed{-5}.$$