We have
$$|m - a_1| |m - a_2| \cdots |m - a_{2n}| = (n!)^2.$$
First we show that we cannot have distinct $i$, $j$, $k$ so that $|m - a_i| = |m - a_j| = |m - a_k|$. If so, then two of $(m - a_i)$, $(m - a_j)$, $(m - a_k)$ must be of the same sign, say $(m - a_i)$, $(m - a_j)$. Then $a_i = a_j$, a contradiction. Thus the values of $|m - a_1|$, $|m - a_2|$, ..., $|m - a_{2n}|$ are $1$, $1$, $2$, $2$, ..., $n$, $n$. Also if $|m - a_i| = |m - a_j|$, then we must have $m - a_i = -(m - a_j)$ otherwise $a_i = a_j$. Therefore $\sum (m - a_i) = 0$ and $m = \frac{a_1 + \cdots + a_{2n}}{2n}$.