(6) Find the smallest positive integer $n$ not less than 9, such that: for any given $n$ integers (which may be the same) $a_{1}, a_{2}, \cdots, a_{n}$, there always exist 9 numbers $a_{i_{1}}, a_{i_{2}}, \cdots, a_{i_{9}}$ $\left(1 \leqslant i_{1}<i_{2}<\cdots<i_{9} \leqslant n\right)$ and $b_{i} \in\{4,7\}(i=1,2, \cdots, 9)$, such that $b_{1} a_{i_{1}}+b_{2} a_{i_{2}}+\cdots+b_{9} a_{i_{9}}$ is a multiple of 9. (Proposed by Yonggao Chen)
Let $a_{1}=a_{2}=1, a_{3}=a_{4}=3, a_{5}=\cdots=a_{12}=0$, then any 9 of these numbers do not meet the requirement, so $n \geqslant 13$. Next, we prove that $n=13$ is possible. For this, it is sufficient to prove that if $m$ integers (which can be the same) $a_{1}, a_{2}, \cdots, a_{m}$ do not contain 3 numbers $a_{i_{1}}, a_{i_{2}}, a_{i_{3}}$ and $b_{1}, b_{2}, b_{3} \in\{4,7\}$ such that $b_{1} a_{i_{1}}+b_{2} a_{i_{2}}+b_{3} a_{i_{3}}$ is a multiple of 9, then $m \leqslant 6$ or $7 \leqslant m \leqslant 8$ and $a_{1}, a_{2}, \cdots, a_{m}$ contain 6 numbers $a_{i_{1}}, a_{i_{2}}, \cdots, a_{i_{6}}$ and $b_{1}, b_{2}, \cdots, b_{6} \in\{4,7\}$ such that $9 \mid b_{1} a_{i_{1}}+b_{2} a_{i_{2}}+\cdots+b_{6} a_{i_{6}}$.
Let
$$
\begin{array}{l}
A_{1}=\left\{i|1 \leqslant i \leqslant m, 9| a_{i}\right\}, \\
A_{2}=\left\{i \mid 1 \leqslant i \leqslant m, a_{i} \equiv 3(\bmod 9)\right\}, \\
A_{3}=\left\{i \mid 1 \leqslant i \leqslant m, a_{i} \equiv 6(\bmod 9)\right\}, \\
A_{4}=\left\{i \mid 1 \leqslant i \leqslant m, a_{i} \equiv 1(\bmod 3)\right\}, \\
A_{5}=\left\{i \mid 1 \leqslant i \leqslant m, a_{i} \equiv 2(\bmod 3)\right\},
\end{array}
$$

then $\left|A_{1}\right|+\left|A_{2}\right|+\left|A_{3}\right|+\left|A_{4}\right|+\left|A_{5}\right|=m$ and:
(1) If $i \in A_{2}, j \in A_{3}$, then $9 \mid 4 a_{i}+4 a_{j}$;
(2) If $i \in A_{4}, j \in A_{5}$, then 9 divides $4 a_{i}+4 a_{j}$, $4 a_{i}+7 a_{j}$, or $7 a_{i}+4 a_{j}$. This is because $4 a_{i}+4 a_{j}$, $4 a_{i}+7 a_{j}$, and $7 a_{i}+4 a_{j}$ are all multiples of 3 and are pairwise distinct modulo 9;
(3) If $i, j, k \in A_{2}$ or $i, j, k \in A_{3}$, then $9 \mid 4 a_{i}+4 a_{j}+4 a_{k}$;
(4) If $i, j, k \in A_{4}$ or $i, j, k \in A_{5}$, then 9 divides $4 a_{i}+4 a_{j}+4 a_{k}$, $4 a_{i}+4 a_{j}+7 a_{k}$, or $4 a_{i}+7 a_{j}+7 a_{k}$. This is because these three numbers are all multiples of 3 and are pairwise distinct modulo 9.
By assumption, $\left|A_{i}\right| \leqslant 2(1 \leqslant i \leqslant 5)$.
If $\left|A_{1}\right| \geqslant 1$, then $\left|A_{2}\right|+\left|A_{3}\right| \leqslant 2$, $\left|A_{4}\right|+\left|A_{5}\right| \leqslant 2$. Thus, $m=\left|A_{1}\right|+\left|A_{2}\right|+\left|A_{3}\right|+\left|A_{4}\right|+\left|A_{5}\right| \leqslant 6$.

Assume $\left|A_{1}\right|=0, m \geqslant 7$, then $7 \leqslant m=\left|A_{1}\right|+\left|A_{2}\right|+\left|A_{3}\right|+\left|A_{4}\right|+\left|A_{5}\right| \leqslant 8$.

Therefore,
$$
\min \left\{\left|A_{2}\right|,\left|A_{3}\right|\right\}+\min \left\{\left|A_{4}\right|,\left|A_{5}\right|\right\} \geqslant 3 .
$$

By (1) and (2), there exist $i_{1}, i_{2}, \cdots, i_{6} \in A_{2} \cup A_{3} \cup A_{4} \cup A_{5}, i_{1}<i_{2}<\cdots<i_{6}$ and $b_{1}, b_{2}, \cdots, b_{6} \in\{4,7\}$ such that $9 \mid b_{1} a_{i_{1}}+b_{2} a_{i_{2}}+\cdots+b_{6} a_{i_{6}}$. In summary, the smallest $n=13$.