Let $ a = f(1) $ and $ b = f(2) $. Then \begin{align*} 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{align*}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 \[f(20152015) = f(1) = f(2) - f(3) = 15 - 20 = \boxed{-5}.\]