First prove that the largest power of $2$ among the numbers $a_i$, $a_{i+1}$, $\dots$, $a_j$ is unique. Let $2^s$ be the largest of the numbers $a_i$, $a_{i+1}$, $\dots$, $a_j$. If there were $k$ and $l$ with $i \le k < l \le j$ such that $a_k = a_l = 2^s$, then they must be of the form $k = 2^s u$ and $l = 2^s v$, where $u$ and $v$ are odd numbers. Since $k < l$, we have $u < v$ and $u+1 < v$. Since $u+1$ is even, the number $m = 2^s(u+1)$ has a divisor $2^{s+1}$, and $k < m < l$, which contradicts the choice of $s$. Thus the largest power of $2$ appears only once among the numbers $a_i$, $a_{i+1}$, $\dots$, $a_j$.

Converting the fractions to the common denominator, the fraction with the largest denominator gives $1$ in the numerator, all others give a positive power of $2$, i.e. an even number. Consequently the numerator is odd and cannot cancel with the denominator.