Proof that Balanced Partition is NP-complete

Instead of adding $k$ zeros for $k$ in $0,1,\ldots,n-1$, just adding $n$ zeros is enough. It is somewhat funny the reduction in the question is just one-off the right reduction.

Here is the detail. It is quite easy.

Let us reduce the partition problem, which is the task of deciding whether a given multiset of positive integers can be partitioned into two subsets with equal sum, to this "balanced partition problem".

Suppose $I=\text{multset}\{a_1, a_2, \cdots, a_n\}$, $a_i>0$ is (the input of) an instance of the partition problem.

Let $I'=\text{multset}\{a_1, a_2, \cdots, a_n, a_{n+1}=0, a_{n+2}=0, \cdots, a_{2n}=0\}$, i.e., $I$ with $n$ zeros added. $I'$ is an instance of the balanced partition problem.

• Suppose $I$ is a yes instance, i.e., there exists $I\subseteq\{1,2,\cdots, n\}$ such that $$\sum_{i\in I}a_i=\sum_{1\le i\le n,\,i\notin I}a_i.$$ Then $$\sum_{i\in J}a_i=\sum_{1\le i\le 2n,\,i\notin J}a_i,$$ where $J=I\cup\{i\mid n+1\le i\le 2n-|I|\}$. Since $|J|=2n/2$, we know $I'$ is a yes instance.
• Suppose $I'$ is a yes instance, i.e., there exists $J\subseteq\{1,2,\cdots, 2n\}$ such that $|J|=n$ and $$\sum_{i\in J}a_i=\sum_{1\le i\le 2n,\,i\notin J}a_i.$$ Removing all items that are $0$ from both sides, we obtain an equality that says $I$ is a yes instance.

The mapping from $I$ to $I'$ is a polynomial-time many-one reduction (a.k.a Karp reduction) from the partition problem to the balanced partition problem. (In fact, it is one-to-one reduction; however, "many-one" is good enough.) Since the former is $\mathsf{NP}$-complete and the latter is in $\mathsf{P}$, the latter is $\mathsf{NP}$-complete as well.