Let $[a_1 a_2 \cdots a_k]_p$ be a number base $p$ with $k$ digits. Multiply each number in $M$ by $7^4$, and we get
$$M' = \{a_1 7^3 + a_2 7^2 + a_3 7 + a_4; a_i \in T, i = 1, 2, 3, 4\} = \{[a_1 a_2 a_3 a_4]'_7 \mid a_i \in T, i = 1, 2, 3, 4\}.$$
The maximum number in $M'$ is $[6666]_7 = [2400]_{10}$.
In the decimal system, starting from 2 400 in the descending order, the 2 005-th number is $2400 - 2004 = 396$. But $[396]_{10} = [1104]_7$. Divide this number by $7^4$, we get a number in $M$, that is $\frac{1}{7} + \frac{1}{7^2} + \frac{0}{7^3} + \frac{4}{7^4}$. Answer: C.