jmc

Let the roots of the quadratic be $r$ and $s$. By Vieta's Formulas, $r+s=-b$ and $rs$ = $2008b$. We know that one of the possible values of $b$ is 0 because $x^2$ has integer roots. However, adding or removing 0 does not affect the value of $S$, so we can divide both sides by $-b$. Doing so results in$$\begin{array}{rl}\frac{rs}{r+s}& =-2008\\ rs& =-2008r-2008s\\ rs+2008r+2008s& =0\\ (r+2008)(s+2008)& =2008^2.\end{array}$$WLOG, let $|a|\le 2008$ be a factor of $2008^2$, so $r+2008=a$ and $s+2008=\frac{2008^2}{a}$. Thus,$$-r-s=b=-a-\frac{2008^2}{a}+4016.$$Since $a$ can be positive or negative, the positive values cancel with the negative values. The prime factorization of $2008^2$ is $2^6\cdot 251^2$, so there are $\frac{21+2}{2}=11$ positive factors that are less than $2008$. Thus, there are a total of $22$ values of $a$, so the absolute value of the sum of all values of $b$ equals $4016\cdot 22=\boxed{88352}$.