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. 1 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 that each digit$d \in [9]$occurs in at most one of the (multi-sets)$S_1$or$S_2$. We claim that no digit occurs more that 54 times, from which the lemma clearly follows. Towards a contradiction suppose that some$i \in [9]$occurs at least 55 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
$$\sum S_1 \geq 55i = \sum_{j=1}^{10} ij > \sum S_2,$$