TITLE: Summable enumerations of $\Bbb Q$
QUESTION [12 upvotes]: We say that a set of natural numbers $A$ is summable if $\sum_{n\in A}\frac1n$ is finite. It is not hard to see that $\{A\subseteq\Bbb N\mid A\text{ is summable}\}$ is an ideal on $\Bbb N$:

Subsets of summable sets are summable.
The union of two summable sets is summable.
The intersection of any number of summable sets is summable.
Every finite set is summable.
$\Bbb N$ is not summable.

Given an enumeration of $\Bbb Q$, we say that it is summable at $q$, or that $q$ is a summable point for the enumeration, if $\{n\mid q_n<q\}$ is summable. And the enumeration is summable if it is summable at each rational point.
It is not hard to see that if $A$ is an infinite summable set, then we can pick any $q\in\Bbb Q$ and we can find an enumeration of $\Bbb Q$ such that $A=\{n\mid q_n<q\}$ (simply take a bijection between $A$ and $\{p\in\Bbb Q\mid p<q\}$ and a bijection of $\Bbb N\setminus A$ with $\{p\in\Bbb Q\mid q\leq p\}$ as the enumeration). So each rational is a summable point for the enumeration.

Question. Is there a summable enumeration of $\Bbb Q$?

This can be translated into a slightly more order theoretic question about the ideal of summable sets:

Question (Reinterpreted). Is there an infinite partition of $\Bbb N$ into infinite summable sets?

If the answer is positive, then by taking such partition into $A_n$'s we can enumerate the negative rationals by $A_0$, then the rationals in the interval $[n-1,n)$ by $A_n$ for $n>0$. It is not hard to see that this enumeration is summable; and vice versa if we are given a summable enumeration, this method produces an infinite partition of $\Bbb N$ into infinite summable enumerations.

REPLY [3 votes]: A very nice question (as can be seen from the large number and diversity of the answers already given)! There's another solution that came to my mind, but it looks so obvious to me that I wonder why nobody has posted it yet. Did I get anything wrong in the definition? Anyway, here it is:
For a nonnegative integer $n\in\mathbb{N}_0$, let $P_n=\{(2n+1)\cdot2^k\,|\,k\in\mathbb{N}_0\}$. Then the sets $P_0$, $P_1$, ... form an infinite partition of $\mathbb{N}$ into infinite summable sets.