At this answer it is provided an interval of the form $[kn,(k+1)n]$ where each integer is divisible by at least one of the integers of the interval $[2,n]$.
We define a permutation of the integers of the interval $[2,n]$ dividing the integers of the interval $[kn,(k+1)n]$ as the set of greatest divisors of $kn,kn+1,...,(k+1)n$ that are also integers of $[2,n]$.
Question 1. I am interested in determining if there are some restrictions (modular or other kind of) in the possible permutations of the odd integers of the interval $[2,n]$ dividing the integers of the interval $[kn,(k+1)n]$.
As there are $n-1$ integers in the interval $[2,n]$ and $n+1$ integers in the interval $[kn, (k+1)n]$, by the Pigeonhole Principle, apart from $kn$ and $(k+1)n$, which are both divisible by $n$, there are at least another two (odd) integers of the interval divisible by the same integer of the interval $[2,n]$; I believe that this "forces" some "well-ordered" permutations of the odd integers, specifically an order similar to that of interval $[2,n]$, that is, in consecutive or reverse-consecutive order.
For instance, at the answer cited, two examples of intervals of the form $[kn,(k+1)n]$ where all integers are divisible by some integer of the interval $[2,n]$ are given for $n=8$: $[200,208]$ and $[632, 640]$. In the first one, the permutation of divisors from interval $[2,8]$ of each integer from interval $[200,208]$ is $\{8,3,2,7,6,5,4,3,8\}$; we have $7,5,3$, the odd integers of $[2,8]$, in reverse-consecutive order. The permutation of divisors for the second one can be seen in the answer cited, and has $3,5,7$ in consecutive order.
Question 2. Even if some permutation exists where this ("forced"?) "well-ordering" is not present (which I have not found), can we nevertheless assure that the minimum solution of an interval of the form $[kn,(k+1)n]$ where each integer is divisible by at least one of the integers of the interval $[2,n]$ is going to be a "well-ordered" permutation?
Thanks for your time!
EDIT
After checking with a computer program, I have found at least one exception. For $n=23$ the minimum solution can be found for $k=58$ and has the following permutation:
\begin{align*} 1334 & : 23 \\ 1335 & : 15 \\ 1336 & : 8 \\ 1337 & : 7 \\ 1338 & : 6 \\ 1339 & : 13 \\ 1340 & : 20 \\ 1341 & : 9 \\ 1342 & : 22 \\ 1343 & : 17 \\ 1344 & : 21 \\ 1345 & : 5 \\ 1346 & : 2 \\ 1347 & : 3 \\ 1348 & : 4 \\ 1349 & : 19 \\ 1350 & : 18 \\ 1351 & : 7 \\ 1352 & : 13 \\ 1353 & : 11 \\ 1354 & : 2 \\ 1355 & : 5 \\ 1356 & : 12 \\ 1357 & : 23 \\ \end{align*}
Nevertheless, I still wonder if this is a rare exception or not.
