Solution:
Suppose first that $a d \neq b c$. The set of points $(a x + b y, c x + d y)$, $x, y \in (0,1)$, coincides with the interior of the parallelogram with vertices $A = (0, 0)$, $B = (a, c)$, $C = (b, d)$ and $D = (a + b, c + d)$. Its area $S$ equals $|a d - b c|$. The Pick formula implies that $S = n + \frac{m}{2} - 1$, where $n$ (respectively $m$) denotes the number of lattice points (i.e., with integer coordinates) in the interior (respectively on the boundary) of the parallelogram. Set $e = (a, c)$, $f = (b, d)$, $a = e a_1$, $c = e c_1$, $b = f b_1$ and $d = f d_1$. The interior points of the side $A B$ have coordinates $(a x, c x)$, $x \in (0,1)$. Thus the number of such lattice points is $e - 1$. Analogously, the number of the interior lattice points on $B D$, $C D$ and $A C$ equals $f - 1$, $e - 1$ and $f - 1$, respectively. Consequently $\frac{m}{2} = e + f$ and the first condition of the problem can be written as
$$e f\left|a_1 d_1 - b_1 c_1\right| = 2003 + e + f$$
Since $e = (a, c) = 6$, it follows that $f$ divides $2009 = 7^2 \cdot 41$ and $6 f$ divides $2009 + f$. This is possible only for $f = 1, 7, 49$. For each of these values of $f$ the numbers $e = 6$, $a_1 = 1 + \frac{2009 + f}{6 f}$, $b_1 = c_1 = d_1 = 1$ satisfy (1) and hence $(b, d) = 1, 7$ or $49$.

Suppose now that $a d = b c$. It is easy to see that $a_1 = b_1$ and $c_1 = d_1$. For $x \in \left(0, \frac{1}{e}\right)$ set $y = \frac{1 - x e}{f} < \frac{1}{f}$. Then $a x + b y = a_1$ and $c x + d y = c_1$ are integers, and $e x + f y = 1$. It follows that in this case there are infinitely many pairs $(x, y)$ satisfying (1), a contradiction.