jmc

In base-$2$ representation, all positive numbers have a leftmost digit of $1$. Thus there are $\left(\genfrac{}{}{0ex}{}{n}{k}\right)$ numbers that have $n+1$ digits in base $2$ notation, with $k+1$ of the digits being $1$'s. In order for there to be more $1$'s than $0$'s, we must have $k+1>\frac{d+1}{2}⟹k>\frac{d-1}{2}⟹k\ge \frac{d}{2}$. Therefore, the number of such numbers corresponds to the sum of all numbers on or to the right of the vertical line of symmetry in Pascal's Triangle, from rows $0$ to $10$ (as $2003<2^{11}-1$). Since the sum of the elements of the $r$th row is $2^r$, it follows that the sum of all elements in rows $0$ through $10$ is $2^0+2^1+\cdots +2^{10}=2^{11}-1=2047$. The center elements are in the form $\left(\genfrac{}{}{0ex}{}{2i}{i}\right)$, so the sum of these elements is ${\sum }_{i=0}^5\left(\genfrac{}{}{0ex}{}{2i}{i}\right)=1+2+6+20+70+252=351$. The sum of the elements on or to the right of the line of symmetry is thus $\frac{2047+351}{2}=1199$. However, we also counted the $44$ numbers from $2004$ to $2^{11}-1=2047$. Indeed, all of these numbers have at least $6$ $1$'s in their base-$2$ representation, as all of them are greater than $1984=11111000000_2$, which has $5$ $1$'s. Therefore, our answer is $1199-44=1155$, and the remainder is $\boxed{155}$.