Let $a_1 = a_2 = 1$, $a_3 = a_4 = 3$, $a_5 = \dots = a_{12} = 0$.
It is easy to check that any 9 integers of them will not meet the requirement. So $n \ge 13$. We will prove that $n = 13$.

We only need to prove the following statement:
Given a group of $m$ integers $a_1, a_2, \dots, a_m$, if there are not three $a_{i_1}, a_{i_2}, a_{i_3}$ in them and $b_1, b_2, b_3 \in \{4, 7\}$ such that $b_1 a_{i_1} + b_2 a_{i_2} + b_3 a_{i_3} \equiv 0 \pmod{9}$, then either $m \le 6$ or $7 \le m \le 8$ and there are $a_{i_1}, a_{i_2}, \dots, a_{i_6}$ in $a_1, a_2, \dots, a_m$ and $b_1, b_2, \dots, b_6 \in \{4, 7\}$ such that $9 \mid b_1 a_{i_1} + b_2 a_{i_2} + \dots + b_6 a_{i_6}$.

We define
$$A_1 = \{i \mid 1 \le i \le m, 9 \mid a_i\},$$
$$A_2 = \{i \mid 1 \le i \le m, a_i \equiv 3 \pmod{9}\},$$
$$A_3 = \{i \mid 1 \le i \le m, a_i \equiv 6 \pmod{9}\},$$
$$A_4 = \{i \mid 1 \le i \le m, a_i \equiv 1 \pmod{3}\},$$
$$A_5 = \{i \mid 1 \le i \le m, a_i \equiv 2 \pmod{3}\}.$$
Then $|A_1| + |A_2| + |A_3| + |A_4| + |A_5| = m$ and

(1) if $i \in A_2, j \in A_3$ then $9 \mid 4a_i + 4a_j$;

(2) if $i \in A_4, j \in A_5$ then one of $4a_i + 4a_j, 4a_i + 7a_j$ and $7a_i + 4a_j$ is a multiple of 9 as all of them are divisible by 3 and they are distinct according to mod 9;

(3) if either $i, j, k \in A_2$ or $i, j, k \in A_3$ then $9 \mid 4a_i + 4a_j + 4a_k$;

(4) if either $i, j, k \in A_4$ or $i, j, k \in A_5$ then one of $4a_i + 4a_j + 4a_k, 4a_i + 4a_j + 7a_k$ and $4a_i + 7a_j + 7a_k$ is a multiple of 9 as all of them are divisible by 3 and they are distinctive according to mod 9.

By the assumption, we have $|A_i| \le 2$ ($1 \le i \le 5$).

If $|A_1| \ge 1$, then $|A_2| + |A_3| \le 2$, $|A_4| + |A_5| \le 2$. Hence
$$m = |A_1| + |A_2| + |A_3| + |A_4| + |A_5| \le 6.$$
Now assume $|A_1| = 0$ and $m \ge 7$. Then
$$7 \le m = |A_1| + |A_2| + |A_3| + |A_4| + |A_5| \le 8.$$
Further,
$$\min\{|A_2|, |A_3|\} + \min\{|A_4|, |A_5|\} \ge 3.$$
From (1) and (2) we know there exist $i_1, i_2, \dots, i_6 \in A_2 \cup A_3 \cup A_4 \cup A_5$ ($i_1 < i_2 < \dots < i_6$) and $b_1, b_2, \dots, b_6 \in \{4, 7\}$ such that $9 \mid b_1a_{i_1} + b_2a_{i_2} + \dots + b_6a_{i_6}$.
The proof of statement is complete.

Now, when $n \ge 13$ it is easy to verify with the statement, for any group of integers $a_1, a_2, \dots, a_n$, there always exist $a_{i_1}, a_{i_2}, \dots, a_{i_9}$ ($1 \le i_1 < i_2 < \dots < i_9 \le n$) and $b_i \in \{4, 7\}$ ($i = 1, 2, \dots, 9$) such that $b_1a_{i_1} + b_2a_{i_2} + \dots + b_9a_{i_9}$ is a multiple of 9. That completes the proof.