Solution:
$4^{2004} + 2^{2004}$

After $n$ successive subdivisions, let $a_{n}$ be the number of small L's in the same orientation as the original one; let $b_{n}$ be the number of small L's that have this orientation rotated counterclockwise $90^{\circ}$; let $c_{n}$ be the number of small L's that are rotated $180^{\circ}$; and let $d_{n}$ be the number of small L's that are rotated $270^{\circ}$. When an L is subdivided, it produces two smaller L's of the same orientation, one of each of the neighboring orientations, and none of the opposite orientation. Therefore,
$$(a_{n+1}, b_{n+1}, c_{n+1}, d_{n+1}) = (d_{n} + 2a_{n} + b_{n},\ a_{n} + 2b_{n} + c_{n},\ b_{n} + 2c_{n} + d_{n},\ c_{n} + 2d_{n} + a_{n})$$
It is now straightforward to show by induction that
$$(a_{n}, b_{n}, c_{n}, d_{n}) = (4^{n-1} + 2^{n-1},\ 4^{n-1},\ 4^{n-1} - 2^{n-1},\ 4^{n-1})$$
for each $n \geq 1$. In particular, our desired answer is $a_{2005} = 4^{2004} + 2^{2004}$.