Stars of Mathematics Competition

The answer is in the affirmative. To prove this, begin by noticing that the rational numbers $b_i = a_1 a_2 \cdots a_n / a_i^2$, $i = 1, 2, \dots, n$, are the roots of the degree $n$ monic polynomial $f = X^n - s_1 X^{n-1} + s_2 X^{n-2} - \dots + (-1)^{n-1} s_{n-1} X + (-1)^n s_n$, where
$$s_k = \sum_{|I|=k} \prod_{i \in I} b_i = \sum_{|I|=k} \left( \prod_{i \in I} a_i \right)^{k-2} \left( \prod_{i \notin I} a_i \right)^k, \quad k = 1, 2, \dots, n.$$
Clearly, $s_2, \dots, s_n$ are all integral. Since $s_1$ is integral, by hypothesis, and $f$ is monic, it follows that the $b_i$ are all integral. (A rational root of a monic polynomial with integral coefficients is necessarily integral.)