Note that $\frac{1}{1+2^{a}}+\frac{1}{1+2^{-a}}=\frac{2+2^{a}+2^{-a}}{2+2^{a}+2^{-a}}=1$. There's $2006$ of these pairs and an additional $\frac{1}{1+2^{0}}=\frac{1}{2}$ so our total sum is $2006+\frac{1}{2}=\boxed{\frac{4013}{2}}$