This isn't really an answer, but just a short proof that there are indeed a finite number of minimal partitions. See ARupinski's answer for a definition of minimal partition.
Lemma. There are a finite set of minimal partitions.
Proof. Let $(S_1, S_2)$ be a minimal partition. Note that $0 \notin S_1 \cup S_2$ and We claim that each no digit $d \in [9]$ occurs more than 44 times in at most one either of the (multi-sets) multi-sets $S_1$ or $S_2$. We claim that no digit occurs more that 54 times, S_2$, from which the lemma clearly follows. Towards a contradiction suppose that some $i \in [9]$ occurs at least 55 45 times in $S_1$. Then for each digit $j$, we have that $j$ occurs at most $i-1$ times in $S_2$ (because otherwise we could remove $j$ copies of $i$ from $S_1$ and $i$ copies of $j$ from $S_2$). But now
$$\sum S_1 \geq 55i 45i = \sum_{j=1}^{10} sum_{j=1}^{9} ij > \sum S_2,$$
a contradiction.
