$2,3,4,6$ The values $n=2,3,4,6$ work, as shown by respective examples 1,$2 ; 2,1,3 ; 2,4,1,3 ; 3,6,2,4,1,5$. We shall show that there are no other possibilities. If $n=2 k+1$ is odd, then none of the numbers $k+1, k+2, \ldots, 2 k+1$ can divide any other, so no two of these numbers are adjacent. This is only possible if they occupy the 1st, 3rd, $\ldots,(2 k+1)$th positions in the line, which means every number $\leq k$ is adjacent to two of these and hence divides two of them. But $k$ only divides one of these numbers when $k \geq 2$. Thus no odd $n \geq 5$ works. If $n=2 k$ is even, the numbers $k+1, k+2, \ldots, 2 k$ again must be mutually nonadjacent, but now this means we can have up to two numbers $\leq k$ each of which is adjacent to only one number $>k$, and if there are two such numbers, they must be adjacent. If $k \geq 4$, then each of $k-1, k$ divides only one of the numbers $k+1, \ldots, 2 k$, so $k-1, k$ must be adjacent, but this is impossible. Thus no even $k \geq 8$ works, and we are done.