As first step, let's prove that f(x) + f(1-x) = 1  for any value of x.
We have 

    f(1-x) = by the definition of function f(x) =  = 

           =  =  =  =  = .


    THEREFORE,  f(x) + f(1-x) =  +  =  = 1,

    and the statement is proved.



As the next step, let's write two identical sums in direct and inverse order

    f(1/2016)     + f(2/2016)    + f(3/2016)    + . . . + f(2015/2016)

    f(20125/2016) + f(2014/2016) + f(2013/2016) + . . . + f(1/2016)


and add them. Pairing the addends vertically, we have 2015 pairs of the form   + , 
and the sum in each such a pair equals 1.


So, the doubled sum equals 2015 and the sum itself is  ,  exactly as the problem states.