Graph it, I dare you!

Maximum occurs at mid point of 1 and 2015, i.e. 1008.

Hence,

$\alpha = 1008$

$f(1008) = 2 \times \displaystyle \sum_{r=1}^{1007} r = \boxed{1015056}$

$\sqrt{\alpha + f(\alpha)} = \sqrt{1015056 + 1008} = \sqrt{1016064} = \boxed{1008}$

To all, here's a solution I once contributed to the Purdue University 'Problem of the Week' Spring 2011 Series that shows how f(x) attains its minimum value at the median value of the absolute-value summation.

We can think of this as a piece wise function and consider intervals [0,1],[1,2]...[2014,2015]. We can prove that the solution lies in the interval by showing that choosing a number greater than 2015 or less than 0 yields a value that is greater than f(0) (or something similar).we can hence consider p positive absolute value functions and 2015 - p negative absolute value expressions. To derive v = -p^2 +2px+p-2015x+2031120. This now becomes an optimization problem under the constraint p=<x<=p+1 which is easy to solve.