I'm currently stuck showing $NP$-hardness of a problem of mine.
An instance of my problem (I call it (simple) pairwise $2$-Partitioning) is given by the following:
Given a set of tupels $B=\{(b_1,1),\ldots,(b_n,1)\}$, with $b_i\in\mathbb{N}_{>0}$.
Goal: A partition of $B$ into two subsets $S_1, S_2$ such that each element $b_i$ is exclusively either in $S_1$ or in $S_2$ and $$(\sum_{i\in S_1}b_i)+|S_2|=(\sum_{i\in S_2}b_i)+|S_1|$$
This can be seen in the following way. When $b_i$ is choosen from the pair $(b_i,1)$ to be in $S_1$ the other Element (in this case always $1$) must be in $S_2$.
I assume that this problem is $NP$-hard and that this can be shown by using the "classical" $2$-Partitioning-Problem where the goal is to partition a set $A=\{a_1,\ldots,a_n\}$ (with $a_i\in\mathbb{N}_{>0}$) into subsets $S_1,S_2$ s.t. the sum of these subset equals.
This far I wasn't able to show that the "classical" $2$-Partiting-Problem can be reduced to (simple) pairwise $2$-Partitioning.
My first attempt was to model some arbitrary but fixed instance of the $2$-Partitioning Problem with $A=\{a_1,\ldots,a_n\}$ to an instance of (simple) pairwise 2-Partitioning by constructing $B=\{(a_1-1,1),\ldots,(a_n-1,1)\}$ but somehow wasn't able to show that each $YES$-Instance of $A$ also is some $YES$-Instance of $B$.
