Interesting problem Define $x_i = \frac{a_i}{i}$ instead, and we know that $a_i \in \mathbb{Z}$ from the problem's condition. Claim 01. $x_{i + 1} - x_i \ge \frac{1}{i} - \frac{1}{i + 1}$ Proof. To prove this, notice that since $x_{i + 1} > x_i$, then \[ \frac{a_{i + 1}}{i + 1} > \frac{a_i}{i} \Rightarrow ia_{i + 1} > (i + 1)a_i \Rightarrow ia_{i + 1} \ge (i + 1)a_i + 1 \]Then \begin{align*} x_{i + 1} &= \frac{a_{i + 1}}{i + 1} \\ &= \frac{ia_{i + 1}}{i(i + 1)} \\ &\ge \frac{(i + 1)a_i + 1}{i(i + 1)} \\ &= \frac{1}{i} a_i + \frac{1}{i(i + 1)} \\ &= x_i + \frac{1}{i} - \frac{1}{i + 1} \end{align*}We are done, since \[ x_n - x_m = \sum_{i = m}^{n - 1} (x_{i + 1} - x_i) = \sum_{i = m}^{n - 1} \left( \frac{1}{i} - \frac{1}{i + 1} \right) = \frac{1}{m} - \frac{1}{n}. \]