Romanian Mathematical Olympiad

Rewrite $2 \le \frac{a}{b} \le 5$ as $2a \le 10b \le 5a$. It follows that the sequence $2a, 2a+1, \dots, 5a$ contains 2014 multiples of 10, hence it contains at least 2013 and at most 2015 groups of 10 consecutive numbers. We deduce that $2013 \cdot 10 \le 5a - 2a < 2015 \cdot 10$, which leads to $a \in \{6710, 6711, \dots, 6716\}$. By inspection, we conclude that the required values of $a$ are: 6710, 6712, and 6713.