The constant 2 is sharp, since we note that for $a_i=\frac{1}{i}$, $LHS\rightarrow \sum_{k=1}^n\frac{2}{k+1}$. Hence for large $n$ it gets arbitrarily close to 2. With this "quasi-equality" case in mind, we estimate the LHS with a linear term: $$\left(\frac{k(k+1)}{2}\right)^2\le \left(\frac{1}{a_1}+\frac{1}{a_2}+...+\frac{1}{a_k}\right)\left(a_1\cdot 1^2+a_2\cdot 2^2+...+a_k\cdot k^2\right)$$ $$\frac{k}{a_1^{-1}+a_2^{-1}+...+a_k^{-1}}\le \frac{4}{k(k+1)^2}\sum_{i=1}^k i^2a_i$$ Throw everything together: \[\sum_{k=1}^n \frac{k}{a_1^{-1}+a_2^{-1}+...+a_k^{-1}}\le \sum_{k=1}^n\left(\frac{4}{k(k+1)^2}\sum_{i=1}^k i^2a_i\right)\le \sum_{i=1}^n\left(\sum_{k\ge i} 2i^2a_i\cdot\frac{2}{k(k+1)^2}\right)<\sum_{i=1}^n\left(\sum_{k\ge i} 2i^2a_i\cdot\frac{1}{i^2}\right)=2\sum_{k=1}^n a_k\] The second last step telescopes because $\frac{2}{k(k+1)^2}=\frac{2k}{k^2(k+1^2}<\frac{2k+1}{k^2(k+1)^2}=\frac{1}{k^2}-\frac{1}{(k+1)^2}$.

It was here: http://www.artofproblemsolving.com/community/c6h169630