jmc

Note that for a given integer $a$, where $1\le a\le 2008$,$$x_a+a=\sum _{n=1}^{2008}{x}_n+2009$$Add up the equations for all $a$ to get$$\sum _{n=1}^{2008}{x}_n+\frac{2009\cdot 2008}{2}=2008(\sum _{n=1}^{2008}{x}_n+2009)$$We can substitue $S={\sum }_{n=1}^{2008}{x}_n$ and simplify to make the equation look easier to solve.$$S+2009\cdot 1004=2008S+2009\cdot 2008$$$$-2007S=2009\cdot 1004$$$$S=\frac{2009\cdot 1004}{-2007}$$Thus, $\lfloor |S|\rfloor =\boxed{1005}$.