Let $a_k$ with $1 \le k \le 2n+2$ denote the minimum. Translating the sequence by a constant, we may assume that $a_k = 0$. Moreover, reversing the order of the sequence if necessary, we may assume that $1 \le k \le n+1$. By the minimality of $a_k$, we have $a_{k+1}, \dots, a_{k+n} \ge 1$ and from the distance assumption we have $a_{k+1}, \dots, a_{k+n} \le a_k + n = n$. Thus the pairwise distinct integers $a_{k+1}, \dots, a_{k+n}$ form a permutation of $1, 2, \dots, n$.

Assuming $k \ge 2$ gives a contradiction: $n+1 \le a_{k-1} \le n + a_k = n$, thus $k = 1$. Similarly, $a_{2n+2}$ is the maximum and since all the numbers are distinct integers, we have $a_{2n+2} \ge 2n+1$. Let $a_l = 1$ for $2 \le l \le n+1$. Then $a_{2n+2} \le n + a_{n+l} \le 2n + a_l = 2n+1$, thus $a_{2n+2} = 2n+1$. This completes the solution.