9. If the sum of the digits of a natural number $a$ equals 7, then $a$ is called a "lucky number". Arrange all "lucky numbers" in ascending order as $a_{1}, a_{2}, a_{3}, \cdots$. If $a_{n}=2005$, then $a_{5 n}=$ $\qquad$.
9. Since the number of non-negative integer solutions to the equation $x_{1}+x_{2}+\cdots+x_{k}=m$ is $C_{m+k-1}^{m}$, and the number of non-negative integer solutions where $x_{1} \geqslant 1, x_{i} \geqslant 0(2 \leqslant i \leqslant k)$ is $C_{m+k-2}^{m-1}$, setting $m=7$, we get the number of $k$-digit “lucky numbers” as $p(k)=C_{k+5}^{6}$. Since 2005 is the smallest “lucky number” in the form $\overline{2 a b c}$, and $p(1)=C_{6}^{6}=1, p(2)=C_{7}^{6}=7, p(3)=C_{8}^{6}=28$, for the four-digit “lucky number” $\overline{1} a b c$, its number is the number of solutions to the equation $a+b+c=6$, which is $C_{6+2}^{6}=28$. Therefore, 2005 is the $1+7+28+28+1=65$th “lucky number”. That is, $a_{65}=2005$. Thus, $n=65, 5 n=325$. Also, $p(4)=C_{9}^{6}=84, p(5)=C_{10}^{6}=210$, and $\sum_{k=1}^{5} p(k)=$ 330. The last six five-digit “lucky numbers” from largest to smallest are $a_{330}=70000, a_{329}=61000, a_{328}=60100, a_{327}=60010, a_{326}=$ $60001, a_{325}=52000$. Hence, $a_{5 n}=a_{325}=52000$.