If the sum of the remaining $2n$ elements is always even, regardless of which element is removed, then all the elements must have the same parity. This means $$a_1\equiv a_2\equiv \cdots \equiv a_{2n+1}\pmod{2}$$

Let $a_i^{(2)} = \frac{a_i}{2}$ or $\frac{a_i - 1}{2}$ depending on the parity of $a_i$ where $1 \le i \le n$. Clearly, after removing any $a_i^{(2)}$, the remaining elements can also be equally divided into two groups with equal sum. This means all these $a_i^{(2)}$ have the same parity. This means that $$\begin{array}{ll} &a_1^{(2)}\equiv a_2^{(2)}\equiv \cdots \equiv a_{2n+1}^{(2)}\pmod{2}\\ \implies& a_1\equiv a_2\equiv \cdots \equiv a_{2n+1}\pmod{2^2}\end{array}$$

This process can be repeatedly applied and as a result the following congruence will hold for every positive integer $k$: $$a_1\equiv a_2\equiv\cdots\equiv a_{2n+1}\pmod{2^k}$$

This cannot be true unless these $a_i$ are all equal.