Suppose that $n$ is an odd prime number. Define a finite sequence $( b_{j})$ of length $n-1$ by $b_{2k} =2k-1$, $b_{2k-1} =n+1-2k$ $( k=1,\dotsc ,( n-1) /2)$. Let $g$ be a primitive root modulo $n$. For any $i\in \mathbb{N} \ ( 1\leq i\leq n-1)$, we can take $a_{i} \in [ 1,n-1] \cap \mathbb{N}$ uniquely such that $a_{i} \equiv g^{b_{i}}\bmod n$. Set $a_{n} =n$. It is not difficult to verify that $( a_{1} ,\dotsc ,a_{n})$ satisfies the condition. $\blacksquare $

If $a_i=n$ for some $i<n$ , then $a_1a_2\dots{a_i}$ and $a_1a_2\dots{a_{i+1}}$ both are divisible by $n$, contradiction . $\Rightarrow{a_n=n}$ . So $a_1a_2\dots{a_{n-1}}=(n-1)!$ and $n$ doesn't divide it by assumption . So $n$ is either a prime number or $n=4$ (well known). If $n=4$ we can take the permutation $a_1=1 , a_2=3 , a_3=2 , a_4=4$ If $n$ is prime , we consider the permutation $a_1=1 , a_n=n$ and $a_i=1+(i-1)^{-1}$ where $(i-1)^{-1}$ is the inverse of $i-1$ modulo $n$ we can easily check that it satisfies the given condition . $\blacksquare$