Summable enumerations of $\Bbb Q$ 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.
 A: Here's another take on the reinterpreted question:
Let $A_0 = \{a^2 : a \in \mathbb{N}\}$ be the set of all perfect squares.  Let $A_1 = \{a + 1: a \in A_0\} - A_0$.  And inductively, define $$A_n = \{a + n: a \in A_0\} - \bigcup_{i < n} A_i$$
So this just shifts the perfect squares, and throws away anything that was already seen.  Since the squares grow arbitrarily long apart, each of these sets will be nonempty (and indeed infinite), and are summable since they are spaced the same as perfect squares.
It's also not hard to see that this is a partition of $\mathbb{N}$.
A: Here's an answer to your reinterpreted question.
Associate to each natural number $m$ except $0$ and $1$ the pair $(p,k)$, such that 


*

*$p$ is the least prime appearing in the factorization of $m$, 

*$m = p^n k$ for some $n$, and

*$p$ does not divide $k$


Let $A_{p,k}$ be the set of natural numbers associated to $(p,k)$. This is nonempty whenever $p$ is prime and all primes dividing $k$ are greater than $p$. Then $A_{p,k} = \{p^nk\mid n\geq 1\}$, which is summable - it's a geometric series.
This gives an infinite partition of $\mathbb{N}\setminus \{0,1\}$ into infinite summable sets. 0 can't be in any summable set, so you'd better leave it out, but you can add $1$ to any of the sets $A_{p,k}$, and it will still be summable.
A: For every natural number $n \ge 1$ define its support $\text{supp}(n)$ to be the set of prime divisors of $n$. We get a surjective map from $\{1,2,3, \ldots,\}$ to the set of finite parts of the set of prime numbers. 
$$\text{supp} \colon \mathbb{N}_{>0} \to \mathcal{P}_{fin}(P)$$
Each nonvoid finite subset of $P$ (the primes ) has an infinite fiber which is clearly summable (Euler's proof of the infinity of the number of primes), with an easy calculable sum ( a product of geometric series).
We got in this way a natural partition of the set of natural numbers $>1$ into a countable family of infinite summable sets. 
A: 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.
A: Let $A$ be some infinite summable set, say, $A=\{1,10,100,\dots\}$.
Partition $A$ into infinitely many disjoint infinite subsets $A_1,A_2,A_3,\dots$.
Let $\mathbb N\setminus A=\{b_1,b_2,b_3,\dots\}$.
Then $\mathbb N$ is the union of the disjoint infinite summable sets
$$A_1\cup\{b_1\},A_2\cup\{b_2\},A_3\cup\{b_3\},\dots.$$
