Summable enumerations of $\Bbb Q$

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
$$A_1\cup\{b_1\},A_2\cup\{b_2\},A_3\cup\{b_3\},\dots.$$

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 $\mathbb{N}$.

Here's an answer to your reinterpreted question.

Associate to each natural number $m$ except $0$ and $1$ the pair $(p,k)$, such that

1. $p$ is the least prime appearing in the factorization of $m$,
2. $m = p^n k$ for some $n$, and
3. $p$ 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.