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Let $x = 20+a$ be a solution to $f(x) = 0$ . If so then $x = 20 - a$ is also a solution as then $f(20-x) = f(20 + x) = 0$ . If we have no solutions where $x = 20$ then there will be an even number of solutions which isn't the case so one of them must be $x = 20$ . The other two solutions must be of the form $20-x, 20+x$ and so the sum of these solutions will be $20 + 20 - x + 20 + x = \boxed{60}$ .

Consider a new coordinate system $O'x'y'$ , which is a translation of the original plane $Oxy$ by a change of variables to $y' = y$ , $x' = x-20$ . This translates the origin from $O(0, 0)$ to $O'(20, 0)$ .

In this system, $f(x+20) = f(x-20)$ becomes $f(x) = f(-x)$ , so $f$ is an even function. Since there are three roots, one of the roots must be at $O'$ and the other two must be equidistant from the origin. Thus, these roots must be $x_0$ and $-x_0$ . Therefore, in the $Oxy$ number plane, these roots are $20$ , $x_0 + 20$ and $-x_0 + 20$ , so their sum is $60$ .

Lets suppose $20$ is not a root. So if there are $0$ roots behind $20$ , f (x) has $0$ roots. If there is $1$ root behind $20$ , f (x) has $2$ roots. If there are $2$ roots behind $20$ , f (x) has $4$ roots and if there are $3$ roots behind $20$ , f (x) has $6$ roots.

So $20$ is a root. Due to symmetry about $x=20$ , we can assume other roots to be $20-a,20+a$

Interestingly, there sum is independent of $a$ , i.e. $60$ .

The given functional equation ensures that that the graph of y=f(x) is symmetric about the line x=20, hence the number of roots of f(x)=0 has to be even. Since it is given to be 3, one of the roots MUST be 20, and the other two, because of the symmetry, will be of the form 20-d and 20+d for some real d. The sum is thus 60.