For every $k \ge 2$ define the quotient $b_k = 2(a_1+\cdots+a_{k-1})/a_k$, which must be a positive integer. We first prove the following properties of the sequence $(b_k)$:

Claim 1. We have $b_{k+1} \le b_k + 1$.

Proof. By subtracting $b_k a_k = 2(a_1 + \cdots + a_{k-1})$ from $b_{k+1} a_{k+1} = 2(a_1 + \cdots + a_k)$, we find that $b_{k+1} a_{k+1} = b_k a_k + 2a_k$. From $a_k < a_{k+1}$ it follows that $b_k + 2 > b_{k+1}$.

Claim 2. The sequence $(b_k)$ is unbounded.

Proof. We start by rewriting $b_{k+1}a_{k+1} = (b_k + 2)a_k$ as
$$a_{k+1} = a_k \cdot \frac{b_k + 2}{b_{k+1}} \Rightarrow a_{k+1} \mid a_k(b_k + 2).$$
If the sequence $(b_k)$ were bounded, say by some positive integer $B$, then the prime factors of the terms of the sequence $(a_k)$ could only be primes less than or equal to $B+2$ or those dividing $a_1$ or $a_2$, which contradicts the property in the statement of the problem.

Consider now an arbitrary positive integer $n$. We assume $n > b_2$, otherwise we replace $n$ by an arbitrary multiple of $n$ that is bigger than $b_2$. By Claim 2, there exists $k$ such that $b_{k+1} \ge n$. Consider the smallest such $k$. From Claim 1, it follows that we must have $b_k = n-1$ and $b_{k+1} = n$ (we assumed $n > b_2$ to ensure that $k \ge 2$). We now find that
$$a_{k+1} = a_k \cdot \frac{b_k + 2}{b_{k+1}} = a_k \cdot \frac{n+1}{n}.$$
Because $n$ and $n+1$ are coprime, this immediately implies that $a_k$ is divisible by $n$.