21.3.11 If the sum of the digits of a positive integer $a$ 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 $a_{n}-2005$, find $a_{5 n}$.
Parse: The number of non-negative integer solutions to the equation $x_{1}+x_{2}+\cdots+x_{k}=m$ is $\mathrm{C}_{m+k-1}^{m}$, and the number of integer solutions where $x_{1} \geqslant 1, x_{i} \geqslant 0(i \geqslant 2)$ is $\mathrm{C}_{m+k-2}^{m-1}$. Taking $m=7$, the number of $k$-digit “lucky numbers” is $P(k)=\mathrm{C}_{k+5}^{6}$.
2005 is the smallest “lucky number” in the form $\overline{2 a b c}$, and $P(1)=\mathrm{C}_{6}^{6}=1, P(2)=\mathrm{C}_{7}^{6}=7, P(3)=\mathrm{C}_{8}^{6}=28$. For four-digit “lucky numbers” $\overline{1 a b c}$, the number of solutions is the number of non-negative integer solutions to $a+b+c=6$, which is $\mathrm{C}_{6+3-1}^{6}=28$.

Since 2005 is the $1+7+28+28+1=65$-th “lucky number”, i.e., $a_{65}=2005$. Thus, $n=65, 5n=325$.
$$
\text{Also, } P(4)=\mathrm{C}_{9}^{6}=84, P(5)=\mathrm{C}_{10}^{6}=210 \text{, and } \sum_{k=1}^{5} P(k)=330 \text{.}
$$

Therefore, the last six five-digit “lucky numbers” in descending order are: $70000, 61000, 60100, 60010, 60001, 52000$. Hence, the 325th “lucky number” is 52000, i.e., $a_{5n}=52000$.