Assume $a_1 \le a_2 \le \cdots \le a_{2n+1}$. By subtracting the smallest number from the set, the remaining set still maintain the the property. Therefore we may assume $a_1=0$. The sum of any $2n$ members must be even, hence any two members must be of the same parity. (otherwise, we may swap members to form two groups of $2n$ elements which are of different parity). Therefore $$0=a_1\equiv a_1 \equiv a_2 \equiv \cdots \equiv a_{2n+1} \pmod{2}$$ Dividing by 2, we note the new sequence will maintain the same property. using the same reasoning, we see that $$0=a_1\equiv a_1 \equiv a_2 \equiv \cdots \equiv a_{2n+1} \pmod{2^2}$$ This process can be repeated infinitely. This is possible only if the initial members are all equal.