The answer is yes. Indeed, for integers $j\ge0$, let
\begin{equation*}
N_j:=\{2^j,\dots,2^{j+1}-1\},\quad
A_j:=2^{-j}\sum_{k\in N_j}a_k,
\end{equation*}
so that $A_j$ is the arithmetic mean of the $a_n$'s over $n\in N_j$.
Then
\begin{equation*}
A:=\sum_{j=0}^\infty 2^jA_j=\sum_{n\ge1}a_n<\infty \tag{1}\label{1}
\end{equation*}
and for $j\ge1$ and $n\in N_j$
\begin{equation*}
\begin{aligned}
s_n:=\sum_{k=1}^n b_k\ge\sum_{k\in N_{j-1}} b_k
&\ge\frac12\, 2^{1-j}\sum_{k\in N_{j-1}} a_k^{-2} \\
&\ge\frac12\,\Big( 2^{1-j}\sum_{k\in N_{j-1}} a_k\Big)^{-2}
=A_{j-1}^{-2}/2;
\end{aligned}
\tag{2}\label{2}
\end{equation*}
the latter inequality here is an instance of Jensen's inequality for the convex function $0<x\mapsto x^{-2}$.
Remark:
Considering the equality case in this latter inequality suggests that the least favorable case is when the $a_n$'s with $n\in N_j$ differ insignificantly from one another (see the explanation "by \eqref{2}" in the multi-line display below) -- but then the $a_n$'s are relatively easy to deal with. Also, the factor $\frac1n$ in the definition $b_n:=\frac{1}{na_n^2}$ of $b_n$ varies insignificantly when $n\in N_j$ -- which was a reason to partition the set of all natural numbers into the blocks $N_j$. $\quad\Box$
Now we can write
\begin{equation*}
\begin{alignedat}{2}
\sum_{n\ge2}\frac n{s_n}
&=\sum_{j\ge1}\sum_{n\in N_j}\frac n{s_n} & \\
&\le\sum_{j\ge1}2^j\frac{2^{j+1}}{A_{j-1}^{-2}/2} & \text{(by \eqref{2})} \\
&=\sum_{j\ge1}2^{2j+2}A_{j-1}^2 & \\
&\le \sum_{j\ge1}2^{2j+2}2^{-(j-1)}A\,A_{j-1} & \text{(by \eqref{1})} \\
&= 16A\sum_{j\ge1}2^{j-1}\,A_{j-1}
=16A^2, &
\end{alignedat}
\end{equation*}
so that the sum of the series in question is
\begin{equation*}
\frac1{s_1}+\sum_{n\ge2}\frac n{s_n}\le\frac1{s_1}+16A^2=a_1^2+16A^2<\infty. \quad\Box
\end{equation*}
