I will show that even the product of the first $n-1$ terms of the set is not divisible by the $n$'th term,term; this is clearly enough, since it has to go to one of the subsets. If $n = 1, 2, \ldots , 6$, then you can check it directly. So, assume that $n > 6$.
By the Zsigmondy's theoremTheorem $10^n - 1$ has a prime divisor $p$ whichthat does not divide $10^k - 1$ for all $k < n$. I claim that $p > 7$ and that $p$ divides $77\ldots 7$ ($n$ sevens). Indeed, $p$ is clearly not $2$ and not $5$, since $10^n-1$ is coprime to $2$ and $5$. We also know that $p$ is not $3$ and not $7$ because $10^6-1$ is divisible by $3$ and $7$, and we assumed that $n > 6$. So, $p > 7$.
That $p$ divides $77\ldots 7$ ($n$ sevens) is also clear now, because it is $7\frac{10^n-1}{9}$ and $p > 7$, so it still divides this number. Similarly, it does not divide any of the numbers $77\ldots 7$ (with $k$ sevens), because, once again, $p > 7$. So, our number does not divide the product of all other numbers, because it is divisible by $p$, and all of the other ones aren't.
