$$\frac{x_1}{x_1+1} = \frac{x_2}{x_2+3} = \frac{x_3}{x_3+5} = ...= \frac{x_{1006}}{x_{1006}+2011}$$$$\frac{x_1+1}{x_1} = \frac{x_2+3}{x_2} = \frac{x_3+5}{x_3} = ...= \frac{x_{1006}+2011}{x_{1006}}$$$$\frac{1}{x_1} = \frac{3}{x_2} = \frac{5}{x_3} = ...= \frac{2011}{x_{1006}}$$$$\frac{x_1}{1} = \frac{x_2}{3} = \frac{x_3}{5} = ...= \frac{x_{1006}}{2011}=k$$Thus, $k(1+3+5+...+2011)=503^2 \implies k=\frac{1}{4} \implies x_{1006}=\frac{2011}{4}$