Each root $r$ in $(\mathbb{Z}/n\mathbb{Z})^\times$ has a ``conjugate'' root $-r \equiv n - r$ since trivially $(-r)^2 \equiv r^2 \pmod{n}$.

Since there are exactly four roots (each prime factor brings in two roots, well, one root and its conjugate) we have exactly two pairs of conjugate roots. In each pair exactly one root will be greater than $n/2$.