We claim the answer is 4. First we show the following claim: Claim: If we have a configuration of numbers 1 through n that works, then removing n leaves a configuration of 1 through n-1 that works. Proof: Focus on the portion of the circle where n is, and the 2 numbers in either side: a b n c d Clearly b+c<2n so b+c=n. Note that b|a+n and c|d+n. When we remove n the only 2 new things we need to check to see if the new configuration works is b|a+c and c|b+d. But clearly these are true as b|a+n-b=a+c and similarly c|d+n-c=d+b. Corollary: all n configurations that work can be formed by adding n to an n-1 configuration that works. This inspires us to use induction. By bashing the n=5 case it is easy to see 12345 12534 and the reverse orders are the only solutions. Now, we claim this: Claim: For all even n at least 6 the only solution is 1 through n in order and reverse, for all odd n at least 5 the only solution is 1 through n in order and 1 through (n-1)/2, n, then (n+1)/2 through n-1, and reverses. Proof: note we must always insert n in between the numbers that sum to n for it to still work. Assume we showed this claim for all n up to k-1, now we show it for k. First, if k is even, then it is easy to check from a parity case that the only solution is 1 through k. And if k is odd, there are obviously exactly two spots to insert k, with its adjacent entries summing to k, which both work. So clearly for all odd n at least 11 the answer is 4.