Given: A circle with nine distinct positive integers. n is a fixed positive integer.
So these positive integers are arranged in the circle in such a way that the product of any selected two NON-ADJACENT numbers in the circle is a multiple of n. The product of two ADJACENT numbers in the circle is NOT a multiple of n. What is the smallest posible value for n.
I'll do (most of) the case of four numbers, and leave the case of nine numbers as an exercise.
I claim that with four numbers, the smallest $n$ is $2\times3\times5\times7=210$.
First, let's see that $210$ is a legitimate value of $n$; let the four numbers, in order around the circle, be $6,15,35,14$. Then the non-adjacent numbers are $6$ and $35$, product $210$, and $15$ and $14$, product $210$, so both divisible by $210$. The adjacent pairs are $6\times15=90$, $15\times35=525$, $35\times14=490$, and $14\times6=84$, none of which is a multiple of $210$.
Now let's show that we can't have $n=pqr$ with $p,q,r$ all prime. If we did, we'd have for each pair of non-adjacent numbers at least one of the number would be divisible by at least two of the primes. Then the numbers adjacent to the two-prime number couldn't be divisible by the third prime. But then their product wouldn't be divisible by the third prime, so not divisible by $n$, contradiction.
Similar reasoning shows we can't have $n=pq$ or $n=p$ for primes $p,q$, so, if $n$ is squarefree, it must have at least four distinct prime divisors, hence, must be at least $210$. So to finish we just have to eliminate the possibility that $n$ has three or fewer distinct prime factors, and is not squarefree.
This case is a bit messier, and I have not written up a complete proof.