Value of sum is $1-\dfrac{1}{4\cdot 2003! +1}$.

Does anybody have solution?

So that we can see the structure, it is given that $\sum_{k=1}^{2002} \dfrac {a_k} {k+n} = \dfrac {4}{2n+1}$ for all $1\leq n\leq 2002$. It is asked to compute $\sum_{k=1}^{2002} \dfrac {a_k} {2k+1}$.

This problem is similar to http://www.artofproblemsolving.com/Forum/viewtopic.php?p=1434406&sid=866a74959481c287b8e5e77966ae6ffe#p1434406

Having $\displaystyle \sum_{k=1}^m \dfrac {a_k}{k+n} = \dfrac {4}{2n+1}$ for each $1\leq n \leq m$, let us consider the polynomial \[p(x) = (2x+1)\left (\prod_{k=1}^m (x+k)\right )\left (\sum_{k=1}^m \dfrac {a_k}{k+x} - \dfrac {4}{2x+1}\right ).\] Since $\deg p \leq m$ and $p(n) = 0$ for all $1\leq n\leq m$, it follows that $\displaystyle p(x) = c\prod_{n=1}^m(x-n)$ for some constant $c$. To compute $c$, let us evaluate $p(x)$ at $x=-1/2$; we get $\displaystyle -4\prod_{k=1}^m (-1/2+k) = p(-1/2) = c\prod_{n=1}^m(-1/2-n)$, whence $\displaystyle c=4(-1)^{m+1}\dfrac {\prod\limits_{k=1}^m (2k-1)} {\prod\limits_{n=1}^m(2n+1)} = 4(-1)^{m+1} \dfrac {1}{2m+1}$. Now write $\displaystyle p(x) = 2(2x+1)\left (\prod_{k=1}^m (x+k)\right )\left (\sum_{k=1}^m \dfrac {a_k}{2k+2x} - \dfrac {2}{2x+1}\right )$ and evaluate $p(x)$ at $x=1/2$, to get $\displaystyle 4\left (\prod_{k=1}^m (1/2+k)\right )\left (\sum_{k=1}^m \dfrac {a_k}{2k+1} - 1\right ) = p(1/2) = 4(-1)^{m+1} \dfrac {1}{2m+1}\prod_{n=1}^m(1/2-n)$, whence $\displaystyle \sum_{k=1}^m \dfrac {a_k}{2k+1} - 1 = -\dfrac {1}{2m+1}\dfrac {\prod\limits_{n=1}^m (2n-1)} {\prod\limits_{k=1}^m(2k+1)} = -\dfrac {1}{(2m+1)^2}$. Thus $\displaystyle \sum_{k=1}^m \dfrac {a_k}{2k+1} = 1 - \dfrac {1}{(2m+1)^2}$. For $m=2002$ that yields $\displaystyle \sum_{k=1}^{2002} \dfrac {a_k}{2k+1} = \boxed{1 - \dfrac {1}{4005^2}}$, not $1 - \dfrac {1}{4\cdot 2003! + 1}$, as claimed in some post above.