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 $f=X^n-s_1{X}^{n-1}+s_2{X}^{n-2}-\cdots +(-1{)}^{n-1}{s}_{n-1}X+(-1{)}^n{s}_n$, where $$s_k=\sum _{|I|=k}\prod _{i\in I}{b}_i=\sum _{|I|=k}{\left(\prod _{i\in I}{a}_i\right)}^{k-2}{\left(\prod _{i\notin I}{a}_i\right)}^k,k=1,2,\dots ,n.$$ 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.)