Theorem: Let $y$ be a quadratic residue in $\mathbb{Z}_N$* where $N=pq$.
There are exactly four integers $x_1, x_2, x_3, x_4$ where $0 < x_1 < x_2 < \frac{N}{2} < x_3 < x_4 < N$ such that $y = x_i^2 \pmod{N}$ for $i=1,2,3,4$.
The above theorem simply states that exactly two of the four roots must be greater than $\frac{N}{2}$.
Most papers will say that this result is well known, without providing any detailed proof. How can we prove that $0 < x_1 < x_2 < \frac{N}{2} < x_3 < x_4 < N$?
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 $r < n/2 \iff n-r > n/2$. 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 $n/2$.
This is not correct for all primes $p,q;\;$ even if $p\ne q:\;$ take e.g. $p=2$, $q=5$. Here you have two quadratic residues in $(\mathbb{Z}/n\mathbb{Z})^\times$ namely $1$ and $9\equiv -1,\;$ but both have only two square roots: $$1^2 \equiv 9^2 \equiv 1 \pmod {10}, \quad \text{and}\quad 3^2 \equiv 7^2 \equiv 9 \pmod {10}$$
