A 2 by 2003 rectangle consists of unit squares as shown below.  The middle unit square of each row is shaded.  If a rectangle from the figure is chosen at random, what is the probability that the rectangle does not include a shaded square?  Express your answer as a common fraction. [asy]
size(7cm);
defaultpen(linewidth(0.7));
dotfactor=4;
int i,j;

fill((6,0)--(7,0)--(7,2)--(6,2)--cycle,gray);

for(i=0;i<=3;++i)

{

draw((i,0)--(i,2));

draw((i+5,0)--(i+5,2));

draw((i+10,0)--(i+10,2));

}
for(j=0;j<=2;++j)

{

draw((0,j)--(3.3,j));

draw((0,j)--(3.3,j));

draw((4.7,j)--(8.3,j));

draw((4.7,j)--(8.3,j));

draw((9.7,j)--(13,j));

draw((9.7,j)--(13,j));

}

real x;

for(x=3.7;x<=4.3;x=x+0.3)

{

dot((x,0));

dot((x,2));

dot((x+5,0));

dot((x+5,2));

}[/asy]
Let $n$ be the number of rectangles contained in the bottom row, and let $m$ be the number of rectangles in the bottom row which contain a shaded square.  There are $n$ rectangles contained in the top row and $n$ rectangles spanning both rows, so there are $3n$ rectangles in the figure.  Similarly, $3m$ rectangles contain a shaded square.  The probability that a rectangle chosen at random includes a shaded square is $3m/3n=m/n$.

A rectangle contained in the bottom row is determined by choosing any two of the 2004 vertical segments as sides of the rectangle.  Therefore, $n=\binom{2004}{2}=\frac{2004\cdot 2003}{2}=1002\cdot2003$.  A rectangle in the bottom row which contains a shaded square is determined by choosing one side from among the 1002 vertical segments to the left of the shaded square and one side from among the 1002 vertical segments to the right of the shaded square.  Therefore, $m=1002^2$.  The probability that a rectangle chosen at random from the figure does not include a shaded square is $1-\dfrac{m}{n}=1-\dfrac{1002^2}{1002\cdot 2003}=1-\dfrac{1002}{2003}=\boxed{\dfrac{1001}{2003}}$.