A simple functional equation

The equation implies that if $(2014-a)$ is a root, $(2014+a)$ must also be a root. However, the only way for there to be an odd number of roots is when $a=0$ , otherwise there will be another distinct root. Thus one of the roots is 2014 and the rest are $(2014-x)$ , and $(2014+x)$ for some real $x$ . Those two roots add up to 4028. Thus the answer is 6042.

The equation implies that if (2014-a) is a root,(2014+a) must also be a root. However, the only way for there to be an odd number of roots is when a=0 , otherwise there will be another distinct root. Thus one of the roots is 2014 and the rest are (2014-x) , and (2014-x) for some real x. Those two roots add up to 4028. Thus the answer is 6042.

the function is symmetric about 2014 and 1 root is 2014

We have $f(x+2014)=f(2014-x)$ $f(x)=f(4028-x)$ So whenever $a$ is a root of $f(x)$ $4028-a$ is also a root of $f(x)$ let b be another root of the function which is not equal to a.This implies (4028-b) is also another distinct root...So the equation will have 4 roots instead of 3 which is a contradiction... So the only possibility is that all the roots must be equal so $a =4028-a$ ...and $a = 2014$

So the sum of all the roots is $2014*3 = 6042$