Notice that if $a_1 < a_2 < \dots < a_n$ are positive integers such that
$a_2/a_1, a_3/a_2, \dots, a_n/a_{n-1} \in (1, 2)$

while $a_n/a_1 \in (n, n+1)$ then $(a_1, \dots, a_n)$ would satisfy the condition of the problem. It thus suffices to find such numbers. We shall then prove the following lemma.

Lemma 1. Let $A$ and $B$ be rational numbers and $p$ and $q$ be distinct prime numbers. Then, $\gcd(A \cdot N! + p, B \cdot N! + q) = 1$, for all large enough positive integers $N$.

Proof. Let $(A, B) = (a/b, c/d)$. Let $D = \gcd(A \cdot N! + p, B \cdot N! + q)$. It follows that $D|acp - bdq$. Thus, $D < \max(acp, bdq)$. Choose $N > 2 \cdot \max(acp, bdq)$, it follows that $D|A \cdot N!$, $D|B \cdot N!$. Yielding $D = 1$. This proves our lemma.

We now construct $a_1, \dots, a_n$. Let us denote by $p_1, \dots, p_n$ the first $n$ primes and $A_1, \dots, A_n$ be arbitrary rational numbers such that $A_i \in (1, 2)$ while $A_1 \cdot A_2 \cdots A_n \in (n, n+1)$. Then, choose a large enough $N$ and put $a_i = A_i \cdot N! + p_i$, $i = 1, \dots, n$. It can be easily verified that these numbers would be satisfying the condition of the problem. ■