12. 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 $a_{5 n}=$ $\qquad$
12.52000 .

It is known that the number of integer solutions to the equation $x_{1}+x_{2}+\cdots+x_{k}=m$ that satisfy $x_{1} \geqslant 1$, $x_{i} \geqslant 0(i \geqslant 2)$ is $\mathrm{C}_{m+k-2}^{m-1}$. Taking $m=7$, we know that the number of $k$-digit lucky numbers is $P(k)=\mathrm{C}_{k+5}^{6}$.

Since 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 the four-digit lucky number $\overline{1 a b c}$, the number of such numbers is the number of non-negative integer solutions to $a+b+c=6$, which is $\mathrm{C}_{6+3-1}^{6}=28$. Therefore, 2005 is the $1+7+28+28+1=65$-th lucky number. Hence, $a_{65}=2005$. Thus, $n=65,5 n=325$.
Furthermore, $P(4)=\mathrm{C}_{9}^{6}=84, P(5)=\mathrm{C}_{10}^{6}=210$, and
$$
\sum_{k=1}^{5} P(k)=330 \text {, }
$$

Therefore, the last six five-digit lucky numbers in descending order are
$$
70000,61000,60100,60010,60001,52000 \text {. }
$$

Thus, the 325th lucky number is 52000.