Reduction from 3-Partition problem to Balanced Partition problem
There are thousands of NP-complete problems in the literature, and most pairs do not have explicit reductions. Since polynomial-time many-one reductions compose, it suffices for researchers to stop when the graph of published reductions is strongly connected, making research into NP-completeness a much more scalable activity.

Although I really don't see the point, I'll humor you by giving a reasonably simple reduction from 3-PARTITION to BALANCED PARTITION, with a few hints about how the proof of correctness goes.

Let the input to the reduction be $x_1, \ldots, x_{3n}, B \in \mathbb Z$, an instance of 3-PARTITION. Verify that $\sum_{i\in[3n]} x_i = nB$. Let $\beta$ be a large number to be chosen later. For every $i \in [3n]$ and every $j \in [n]$, output two numbers
$$x_i \beta^j + \beta^{n+j} + \beta^{2n+i} + \beta^{(i+4)n+j}\\
\beta^{(i+4)n+j}.$$
Intuitively, the first number means that $x_i$ is assigned to 3-partition $j$, and the second number means the opposite. The $x_i \beta^j$ term is used to track the sum of 3-partition $j$. The $\beta^{n+j}$ term is used to track the cardinality of 3-partition $j$. The $\beta^{2n+i}$ term is used to ensure that each $x_i$ is assigned exactly once. The $\beta^{(i+4)n+j}$ term is used to force these numbers into different balanced partitions.

Output two more numbers
$$1 + \sum_{j\in[n]} \Bigl((n-2)B\beta^j + (3n-6)\beta^{n+j}\Bigr) + \sum_{i\in[3n]} (n-2)\beta^{2n+i}\\
1.$$
The first number identifies its balanced partition as &ldquo;true&rdquo;, and the other, as &ldquo;false&rdquo;. The $1$ term is used to force these numbers into different balanced partitions. The other terms make up the difference between the sum of a 3-partition and the sum of its complement and the size of a 3-partition and the size of its complement and the number of times $x_i$ is assigned.

$\beta$ should be chosen large enough to ensure that &ldquo;overflow&rdquo; cannot occur.