jmc

If $$\left|x^2-x+\frac{1}{2008}\right|=\frac{1}{2008},$$then either $x^2-x+\frac{1}{2008}=\frac{1}{2008}$ or $x^2-x+\frac{1}{2008}=-\frac{1}{2008}.$

In the first case, $x^2-x=x(x-1)=0,$ so $x=0$ or $x=1,$ and the sum of the squares is $0^2+1^2=1.$

In the second case, $$x^2-x+\frac{1}{1004}=0.$$Let the roots be $a$ and $b.$ Then by Vieta's formulas, $a+b=1$ and $ab=\frac{1}{1004},$ so $$a^2+b^2=(a+b{)}^2-2ab=1-\frac{1}{502}=\frac{501}{502}.$$Therefore, the sum of the squares of the solutions is $1+\frac{501}{502}=\boxed{\frac{1003}{502}}.$