Convergence of $\sum_{n}\frac{q_n}{n}$, where $(q_n)$ enumerates $\mathbb{Q}\cap[0,1]$?

This problem is from a book reviewed Sep. 2020 Notices of AMS.

Rational numbers in $[0,1]$ are countable. Can they be ordered as $(q_n)_n$ so that $\sum_{n=1}^\infty \frac{q_n}{n}$ converges?

My belief is no, since half the rationals are in the upper half of the interval, the tail of the sum will always have half of its terms $\gt \frac{1}{2n}$ implying divergence. I am having difficulty in making this rigorous.

Ruy makes a good point in the comments you could possibly formalise it as follows.

Consider any ordering of the rational numbers in $[0,1]$ and remove those elements of the form $\frac{1}{n}$ where $n$ is a positive integer which is not a positive power of $2$.

This leaves the sequence $\{a_n\}_{n=1}^\infty$

Then define $$ q_n = \left\{ \begin{array}{ll} a_{\log_2 n}, \text{ if n is a positive power of 2}\\ \frac{1}{n}, \text{ otherwise}\\ \end{array} \right.$$

This will imply that $$ \displaystyle \sum_{n=1}^{\infty} \frac{q_n}{n} < \sum_{n=1}^{\infty} \left( \frac{1}{n^2} + \frac{1}{2^n} \right) = \frac{\pi^2}{6} + 1$$

Here is a proof of the conjecture made by @ClementC, with an improvement suggested by @TSF.

Theorem. Given any bounded function $f:\mathbb N\to \mathbb R_+$ admiting a subsequence $\{f(n_k)\}_k$ that converges to zero, there is an enumeration $\{q_n\}$ of the rationals in $[0,1]$ such that $$ \sum_nq_nf(n)<\infty. $$

Proof. By hypothesis there are arbitrarily small elements in the range of $f$ so we may choose an infinite subset $N_1\subseteq \mathbb N$ such that $\sum_{n\in N_1}f(n)$ converges. By discarding infinitely many elements from $N_1$, if necessary, we may assume that $$ N_2:=\mathbb N\setminus N_1 $$ is also infinite.

Furthermore let $Q_1$ be the set of rationals defined by $$ Q_1=\{1/2^n: n\in \mathbb N\}, $$ and let $Q_2$ be the complement of $Q_1$ in $\mathbb Q \cap [0,1]$.

All sets so far defined are countably infinite, so there are bijections $$ \alpha :N_1\to Q_2, $$ $$ \beta :N_2\to Q_1. $$ The union of $\alpha $ and $\beta $ is therefore a bijection, $$ \gamma :\mathbb N\to \mathbb Q \cap [0,1] $$ which is the enumeration we are looking for, that is, $q_n=\gamma (n)$. We then have that $$ \sum_{n\in \mathbb N} q_nf(n) = \sum_{n\in N_1} \gamma (n)f(n) + \sum_{n\in N_2} \gamma (n)f(n) $$ $$= \sum_{n\in N_1} \alpha (n)f(n) + \sum_{n\in N_2} \beta (n)f(n) \leq \sum_{n\in N_1} f(n) + \|f\|_\infty\sum_{n\in \mathbb N} 1/2^n < \infty. \tag*{$\blacksquare$} $$