The answer is in the affirmative. If $n$ is odd, set $a_1 = 1$ and $a_2 = 2$, and if $n > 3$ define the other $a_k$ recursively by $a_k a_{k-1} = a_0 + a_1 + \dots + a_{k-1}$, $k = 3, \dots, n$. It is easily seen that $a_{k+1} = a_{k-1} + 1$, $k = 3, \dots, n-1$, so $a_{2k} = k+1$ and $a_{2k+1} = k + (n+1)/2$, $k = 1, \dots, (n-1)/2$. By construction, the integers $a_k$ satisfy the divisibility condition (even at $k = n$). The $a_{2k}$ and $a_{2k+1}$, $k = 1, \dots, (n-1)/2$, form strictly increasing sequences of integers greater than 1, and no $a_{2i}$ equals an $a_{2j+1}$, $i, j = 1, \dots, (n-1)/2$, so the $a_k$, $k = 1, \dots, n-1$, form an injective sequence of positive integers. Since they all lie in the range $1, 2, \dots, n-1$, they form indeed a permutation of the latter.

Similarly, if $n$ is even, set $a_1 = 2$ and define $a_k$, $k = 2, \dots, n-2$, by the same recursive relation, to get $a_{2k-1} = k+1$ and $a_{2k} = k + n/2$, $k = 1, \dots, n/2 - 1$. Setting $a_{n-1} = 1$ settles the case as above and completes the proof.

Remark. For an odd $n > 3$, we may equally well start by setting $a_1 = 1$ and $a_2 = (n+1)/2$, and define the remaining $a_k$ by the recursive relation in the solution, to obtain another sequence of positive integers satisfying the divisibility condition. Explicitly, this time $a_{2k} = k + (n-1)/2$, $k = 1, \dots, (n-1)/2$, and $a_{2k+1} = k + 2$,