I have revised my original post. The questions I asked there were not well-put or even thought through. I don't want to delete, however, since some of the comments may be of interest to other MO users.

It is well known that the sums of the reciprocals of the primes diverges. But suppose we decompose the set of prime numbers, call this set $P$, into disjoint infinite sets $\langle A_\alpha\rangle$ in some way to guarantee that each member $A_\alpha$ of the partition has the following "nice" property: $$\Sigma_{p\in A_{\alpha}} \frac{1}{p} \text{ converges}.$$

There are by now several interesting partitions in the comments and answers below. Here I include two of my own attempts at such a partition.

**Attempt 1:** Let $\langle \epsilon_i: i<\omega\rangle$ be a sequence of positive real numbers be such that $\Sigma_i \epsilon_i$ diverges (goes to infinity). Suppose you want an infinite $X\subseteq P$ such that $$\Sigma_{p\in X}\frac{1}{p} <\epsilon_1.$$ Then inductively define $X_1$ as follows: Pick $p_1\in P$ least such that $\frac{1}{p}<\epsilon_1$. Continuing, suppose $p_1,p_2,\dots,p_n$ have been chosen such that $$\Sigma^n_1\frac{1}{p_i}<\epsilon.$$ Then let $p_{n+1}$ be the least such that $$\Sigma^{n+1}_1\frac{1}{p_i}<\epsilon_1.$$

It seems like we could continue this strategy to produce a partition with the desired property by setting $P=P_1$, and $P_2=P_1\setminus X_1$. Then repeat the process described above. If this stage goes all the way through, define $P_3$, etc. It seems that all the $P_n$s will be defined and each will, by construction converge.

But this sort of strategy seems to depend on the paramater sequence $\langle\epsilon_i\rangle$ in an essential way. It may be constructive, but the strategy doesn't really give me a concrete or satisfying (or even really explicit) definition for the partition.

**Attempt 2:** I proposed the following partial strategy (based on the Green-Tao Theorem). Index the primes according to their natural order (i.e., $p_1 =2, p_2 =3, p_3=5$, etc.). Let $P=A_1\cup B_1$ where $$n\in A_1\Longleftrightarrow n \text{ has a prime index},$$ and $n\in B_1$ otherwise. Since $B$ has positive upper density, it must contain arbitrarily long arithmetic progressions by the Tao-Green Theorem (see the related question Extension of Tao-Green Theorem). I'm not sure if this means $$\Sigma_{p\in B_1} \text{ diverges }$$ but I think it does.

Also, Ben Green comments there that $A$ must contain arbitrarily long arithmetic progressions since the density of this set is roughly $1/\log^2(N)$ (though this may still be conjecture, I'm not sure).

But suppose we attempt to destroy each such arithmetic progression by decomposing both $A_1$ and $B_1$ in the following way: Re-index both $A_1$ and $B_1$ according to their natural order and partition each by placing an $n$ in one set of the partition if it is indexed by a prime, and placing it in the other set in the partition if it indexed by a composite.

If we repeat the construction $k$ times (for both $A_1$ and $B_1$ and all subsequently formed subsets) we must surely eliminate all arithmetic progressions with common difference $k$. But perhaps there are still arithmetic progressions with common difference $k'>k$. So we continue the iteration, creating a binary tree of height $\omega$, the leaves of which union to all of $P$.

It is still not clear to me that this partition will have the property I requested above. Certainly there are several nice examples in the comments and answers below.