This 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 rule 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. However the scheme above is easy to follow.