Here is a proof.
We try using induction, but as usual, a direct approach seems to fail and we have to try and prove a stronger statement.
So we try and pick a positive function $f(n)$ such that
$ \frac{f(n)}{a_1 + a_2 + \dots + a_n} + \sum_{k=1}^{n} \frac{k}{a_1 + a_2 + \dots + a_k} \le \sum_{j=1}^{n} \frac{2}{a_j}$
Let $S = a_1 + a_2 + \dots + a_n$ and let $x = a_{n+1}$.
In order to prove that $n$ implies $n+1$ it would be sufficient to prove
$\frac{2}{x} + \frac{f(n)}{S} \ge \frac{f(n+1) + n+1}{S+x}$
This can be rearranged to
$2S^2 + f(n) x^2 + (f(n) + 2) Sx \ge (f(n+1) + n+1) Sx$
Since $ 2S^2 + f(n) x^2 \ge 2 \sqrt{2 f(n)} Sx$
it is sufficient to prove that $f(n)$ satisfies
$ f(n) + 2 + 2\sqrt{2 f(n)} \ge f(n+1) + n + 1$
Choosing $f(n) = \dfrac{n^2}{2}$ does the trick.
We can easily verify the base case for this choice of $f(n)$.
Thus we have:
$ \frac{n^2}{2(a_1 + a_2 + \dots + a_n)} + \sum_{k=1}^{n} \frac{k}{a_1 + a_2 + \dots + a_k} \le \sum_{j=1}^{n} \frac{2}{a_j}$