TW-TST 2002 Problem 5

Suppose that the real numbers $a_{1},a_{2},...,a_{2002}$ satisfying $\frac{a_{1}}{2}+\frac{a_{2}}{3}+...+\frac{a_{2002}}{2003}=\frac{4}{3}$ $\frac{a_{1}}{3}+\frac{a_{2}}{4}+...+\frac{a_{2002}}{2004}=\frac{4}{5}$ $...$ $\frac{a_{1}}{2003}+\frac{a_{2}}{2004}+...+\frac{a_{2002}}{4004}=\frac{4}{4005}$ Evaluate the sum $\frac{a_{1}}{3}+\frac{a_{2}}{5}+...+\frac{a_{2002}}{4005}$ .