Let $M \geq 1$, and $N=2^M$. Let $a_1, \dots, a_N$ be the set of all the numbers that you get when forming all square-free products of the first $M$ primes.

For example, for $M=2$ and $N=4$ you get $\{a_1, a_2, a_3, a_4\}=\{1,2,3,6\}$.

The question is: Find a good upper bound for the sum $$ \sum_{1 \leq m \neq n \leq N} \frac{1}{|\log a_m/a_n|}. $$ I want to find an upper bound for this sum which is polynomial in $N$. (From the maximal size of $a_n$ on can quite easily deduce an upper bound of size roughly $\exp(\log N \log \log N)$, which does not help me. It really has to be polynomial in $N$.)

The problem turns up in the context of proving an upper bound for $$ \int_0^T \sum_{1 \leq m,n \leq N} \cos(t (\log a_m/a_n))~dt, $$ where $T$ is of size $N^\beta$ for some (arbitrary, but fixed) exponent $\beta$. Evidently, the terms for which $n=m$ contribute $TN$. If for a single summand we have $n \neq m$, then the integral is $$ \frac{\sin (T (\log a_m/a_n))}{\log a_m/a_n} \leq \frac{1}{|\log a_m/a_n|}. $$ The question is, whether the $TN$ from the terms with $n=m$ is the dominant contribution, or whether the other terms can be significantly larger. Thus a small number of pairs $m,n$ for which $a_m/a_n$ is very close to 1 is no problem, since the integral is always bounded by $T$. That means, $cN$ terms for some constant $c$ may be discarded from the sum.

The problem is somewhat related to Baker's theorems on linear forms of logarithms, etc. (which is not strong enough to get something here). However, the setting here is quite specific, so maybe someone has a good idea.