ELMO 2015 Problem 2

Let $m$ , $n$ , and $x$ be positive integers. Prove that $$\sum_{i = 1}^n \min\left(\left\lfloor \frac{x}{i} \right\rfloor, m \right) = \sum_{i = 1}^m \min\left(\left\lfloor \frac{x}{i} \right\rfloor, n \right).$$ Proposed by Yang Liu