Reducibility: Show that SUBSET SUM is reducible to the following problem A
You can consider the numbers $x_i$ as signed integers representing how much each person owes (if the number is positive) or how much each person needs to receive (if the number is negative). Then you have $\sum_{i=1}^n x_i = 0$ and the goal is to make all numbers $0$.
Notice that you can always settle debts with at most $n-1$ checks:
pick an $i$ such that $x_i&gt;0$ and move $x_i$ from $i$ to an arbitrary $j \neq i$.
Delete $x_i$ (now $0$) from the instance. You are left with another instance of the same problem in which you need settle debts between $n-1$ people. Repeat recursively until you are left with only one person (in which cases no transactions are needed).
Suppose now that you can partition the set $\{x_1, \dots, x_n\}$ into two non-empty sets $S_1$ and $S_2$ such that $\sum_{x \in S_1} x =0$ (and hence $\sum_{x \in S_2} x = 0$). Then you can solve $S_1$ and $S_2$ separately using at most $|S_1|-1 + |S_2|-1 = n-2$ checks.
The converse is also true: if you can balance debts using at most $n-2$ checks, then you can partition $\{x_1, \dots, x_n\}$ into non-empty $S_1$ and $S_2$ with $\sum_{x \in S_1} x = \sum_{x \in S_2} x = 0$.
To see this, build the graph $G=(V,E)$ in which $V=\{1, \dots, n\}$ and there is an undirected edge $(i,j)$ if $i$ and $j$ exchange at least one check. Since $|E| \le n-2 = |V|-2$, the graph $G$ has at least $2$ connected component. If $C$ is a connected component then the sum of the integers $x_i$ such that $i \in C$ must be $0$. Let $S_1 = \{x_i \mid i \in C\}$ and $S_2 = \{x_1, \dots, x_n\} \setminus S_1$.
This shows that you can reduce the variant of partition in which the sum of the integers is $0$ (and you need to find a non-trivial partition) to your problem.
To see that this variant remains $\mathsf{NP}$-hard, consider an instance of partition $\{y_1, \dots, y_n\}$ in which each integer $y_i$ is positive, and add two additional integers $y'$ and $y''$ with $y'=y''= -\frac{1}{2}\sum_{i=1}^n y_i$.
Clearly, any non-trivial partition $A,B$ of $\{y', y'', y_1, \dots, y_n\}$ must place $y'$ in $A$ and $y''$ in $B$, or vice-versa, showing that the sum of the elements in $A \cap \{ y_1, \dots, y_n\}$ is $\frac{1}{2}\sum_{i=1}^n y_i$ (and the same holds for $B$). Conversely, given any partition $A,B$ of $\{y_1, \dots, y_n\}$, the partition $A \cup \{y'\}$, $B \cup \{y''\}$ is a solution to our variant.