Since the number of non-negative integer solutions of the equation $x_1 + x_2 + \dots + x_k = m$ is $C_{m+k-1}^k$, the number of integer solutions, when $x_1 \ge 1$ and $x_i \ge 0\ (i \ge 2)$, is $C_{m+k-2}^{m-1}$. Let $m=7$, the number of lucky numbers with $k$ digits is $p(k) = C_{k+5}^6$.

Since $2005$ is the minimum lucky number of the type $\overline{2abc}$ and $p(1) = C_6^6 = 1$, $p(2) = C_7^6 = 7$, $p(3) = C_8^6 = 28$. Note that the number of four-digit lucky numbers of the type $\overline{1abc}$ is the number of non-negative integer solutions of $a+b+c=6$, i.e. $C_{6+3-1}^6 = 28$. Thus $1+7+28+28+1=65$ and $2005$ is the $65$-th lucky number, i.e. $a_{65}=2005$, so $n=65$, $5n=325$.

Furthermore $p(4) = C_9^9 = 84$, $p(5) = C_{10}^9 = 210$ and $\sum_{k=1}^{5} p(k) = 330$.

Therefore the last six lucky numbers with $5$ digits, from the largest to the smallest, are $70\ 000$, $61\ 000$, $60\ 100$, $60\ 010$, $60\ 001$, $52\ 000$. So the $325$-th lucky number is $52\ 000$, i.e. $a_{325} = 52\ 000$.