To prove that if there are $2n+1$ positive integers satisfying the property described in the problem, we call these $2n+1$ positive integers a "good array". Let's assume $a_{1} \leqslant a_{2} \leqslant \cdots \leqslant a_{2 n+1}$, and $a_{1}=1$ (because reducing the smallest number to 1 and subtracting the same number from the other numbers still results in a "good array").

From the problem statement, we know that $2 \mid\left(\left(\sum_{i=1}^{2 n+1} a_{i}\right)-a_{j}\right)(j=1,2,3, \cdots, 2 n+1)$, which means $a_{1} \equiv a_{2} \equiv a_{3} \equiv a_{4} \equiv \cdots \equiv a_{2 n+1} \equiv \sum_{i=1}^{2 n+1} a_{i}(\bmod 2)$.

Since $a_{1}=1$, it follows that $a_{1}, a_{2}, a_{3}, \cdots, a_{2 n+1}$ are all positive odd numbers.
Let $b_{i}=\frac{a_{i}+1}{2}(i=1,2, \cdots, 2 n+1)$. Since $a_{1}, a_{2}, \cdots, a_{2 n+1}$ are all positive odd numbers, $b_{1}, b_{2}, b_{3}, \cdots, b_{2 n+1}$ are all positive integers and it is easy to see that $b_{1}, b_{2}, b_{3}, \cdots, b_{2 n+1}$ also form a "good array".

If $x > 1$, then $b_{2 n+1}=\frac{a_{2 n+1}+1}{2}<a_{2 n+1}$. Repeatedly performing this transformation, we will eventually obtain a "good array" whose maximum term is 1. Since in the above transformation, the smallest term in each array is always 1, the terms in this "good array" are all 1. Considering the way the transformation is performed (changing $x$ to $\frac{x+1}{2}$), it follows that the original $2n+1$ numbers $a_{1}, a_{2}, \cdots, a_{2 n+1}$ are all 1.

This completes the proof of the original proposition.