Exactly two of the four roots must be greater than N/2
Each root $r$ in $(\mathbb{Z}/n\mathbb{Z})^\times$ has a ``conjugate'' root $-r \equiv n - r$ since trivially $(-r)^2 \equiv r^2 \pmod{n}$.

If there are exactly four roots (each prime factor generally brings in two roots, well, one root and its conjugate, and they generate the roots modulo $n$ via by CRT - see gammatester's answer below for more details) we have exactly two pairs of conjugate roots. In each pair exactly one root will be greater than $n/2$.

By simple arithmetic, one can see that $r &lt; n/2 \iff n-r &gt; n/2$.  Thus, assuming that $n$ is odd (which rules out the possibility that $r = n/2$), it follows that exactly one of each conjugate pair of roots is grater than $n/2$.