There is nothing special going on related to primes or even to integers (see below.) But let me start with the positive integers. What you want can't be harder than partitioning the positive integers similarly. Consider partitioning the positive integers in sequences as follows:
$\ 1,\ 3,\ 6,10,15,21\cdots$
$\ 2,\ 5,\ 9,14,20,\cdots$
$\ 4,\ 8,13,19,26\cdots$
$\ 7,12,18,25,33,\cdots$
$11,\cdots$
The growing triangular pattern should be clear, but to be specific, row $k$ starts with $1+\frac{k(k-1)}{2}$ and then the gaps between successive terms are $k+1,k+2,k+3,\cdots.$
The sum of the reciprocals in the first row is $2$ and each other row has a smaller sum than the one above it.
Now replace $n$ in this partition with the $n$th prime instead to get what you want.
It would be possible to arrange to have the first row converge quite slowly and then have every following row converge even more slowly (but still converge) and that would work as well. More rapid convergence is possible as well. However the scheme above is specific and easy to follow.
Rather than the sequence of reciprocals of integers or primes, assume merely a sequence of positive reals $ x_1 \ge x_2 \ge \cdots $ such that
- $\lim x_k =0$ and
- $\sum x_k=\infty.$
Suppose we wish to partition into a countable number of infinite subsequences such that the $m$th subsequence has sum $s_m$ where the values $s_1,s_2,\cdots$ are some positive reals. We can't always do this, for example the sum of the $s_m$ has to diverge and we can't have all the $s_m$ less than $\frac{1}{3}$ if $x_3=\frac{1}{2}$.
Here is a sufficient condition:
- There is a sequence $N_1 \lt N_2 \lt \cdots$ such that $x_m \lt s_{N_m}$
So this allows $s_m=x_m+m!$ (huge sums) or $s_m=x_m(1+\frac1{m!})$ (small sums) as well as a mix of extremely small and extremely large target sums. With this condition we can make the subsequences by the greedy rule of considering the $x_i$ in turn and assigning $x_i$ to subsequence $m$ where $m$ is the smallest possible index such that the partial sum will remain strictly less than $s_m.$ This might mean that it takes longer ( at 10 placements per second) than the life of the universe to put anything in subsequence $2$ (or to put the second member into subsequence $1$), but infinity is much larger.
The condition and greedy rule ensure that $x_m$ will be assigned to subsequence $N_m$ or an earlier one. How do we know that each subsequence is infinite and with the proper sum? Pick any $\varepsilon \gt 0.$ There is an $N$ so large that $x_N \lt \varepsilon.$ The $x_i$ for $i \gt N$ will never be assigned past subsequence $m$ until the partial sum for subsequence $k$ is at least $s_k-\varepsilon$ for all $1 \le k \le m.$ However we do at some later stage assign past subsequence $m$ since $\sum_1^ms_i$ is finite but $\sum x_k$ is not.