Taiwanese Mathematical Olympiad 2002

\[\begin{align} \dfrac{a_{1}}{2} + \dfrac{a_{2}}{3} + \cdots + \dfrac{a_{2002}}{2003} &= \dfrac{4}{3}\\\\ \dfrac{a_{1}}{3} + \dfrac{a_{2}}{4} + \cdots + \dfrac{a_{2002}}{2004} &= \dfrac{4}{5}\\ & \vdots\\ \dfrac{a_{1}}{2003} + \dfrac{a_{2}}{2004} + \cdots + \dfrac{a_{2002}}{4004} &= \dfrac{4}{4005} \end{align}\]

Given that real numbers \(a_1, a_2, \ldots, a_{2002}\) satisfy the system of equations above, evaluate the following sum: \[\dfrac{a_1}{3} + \dfrac{a_2}{5} + \cdots + \dfrac{a_{2002}}{4005}.\]

If your answer is of the form \(\dfrac{m}{n}\), where \(m\) and \(n\) are coprime positive integers, submit \(m+n\).