Five, let $\overline{a b c d}$ be a four-digit number, where the first two digits and the last two digits form two-digit numbers $\overline{a b}$ and $\overline{c d}$, respectively. If the following conditions are met: $\overline{a b} \neq \overline{c d}$, $\overline{a b}$ divides $\overline{c d}$, and $\overline{c d}$ divides $\overline{a b c d}$, then such a four-digit number is called a “lucky number”.
(1) Find the total number of all “lucky numbers”;
(2) Find the largest odd “lucky number”.
(1) Since $\overline{a b} \mid \overline{c d}$ and $\overline{a b} \neq \overline{c d}$, we can set $\overline{c d}=k \overline{a b}(k \in \mathbf{N}, k>1)$, and since $\overline{c d}$ is a two-digit number, we know $k<10$, so $2 \leqslant k \leqslant 9(k \in \mathbf{N})$.
Since $\overline{a b c d}=\overline{a b} \times 100+\overline{c d}$ and $\overline{c d} \mid \overline{a b c d}$,
$\therefore \overline{c d} \mid \overline{a b} \times 100$,
i.e., $k \overline{a b}|\overline{a b} \times 100 \Rightarrow k| 100$,
$\therefore \quad k=2,4,5$.
(i) If $k=2$, since $2 \times 50=100, \overline{a b}$ can take any two-digit number from 10 to 49, so there are 40 "lucky numbers" in this case.
(ii) If $k=4$, since $4 \times 25=100, \overline{a b}$ can take any two-digit number from 10 to 24, so there are 15 "lucky numbers" in this case.
(iii) If $k=5$, since $5 \times 20=100, \overline{a b}$ can take any two-digit number from 10 to 19, so there are 10 "lucky numbers" in this case.

Therefore, the total number of "lucky numbers" is $40+15+10=65$.
(2) If $\overline{a b c d}$ is an odd "lucky number", then only $k=5$ is possible, so $\overline{a b}$ can take the maximum value of 19, thus $\overline{c d}=5 \times 19=95$, i.e., the largest odd "lucky number" is 1995.