Solution:

Since $2003 \lt 2^{11} = 2048,$ every integer in question has at most $11$ binary digits. A $(d+1)$-digit binary number starts with $1,$ and choosing $k$ more $1$'s among the remaining $d$ digits gives $\binom{d}{k}$ numbers with $k + 1$ ones; the $1$'s outnumber the $0$'s exactly when $k \ge \frac{d}{2}.$ So the count over all numbers up to $2047$ is the total of the entries on or to the right of the center of rows $0$ through $10$ of Pascal's triangle.

Those rows sum to $1 + 2 + \cdots + 2^{10} = 2047,$ and the central entries sum to $\sum_{i=0}^{5}\binom{2i}{i} = 1 + 2 + 6 + 20 + 70 + 252 = 351,$ so by symmetry the count is $\frac{2047 + 351}{2} = 1199.$

The $44$ integers from $2004$ to $2047$ all exceed $1984 = 11111000000_2,$ so each has the prefix $11111$ plus at least one more $1,$ hence at least six $1$'s among eleven digits — all $44$ were counted. Therefore $N = 1199 - 44 = 1155,$ whose remainder upon division by $1000$ is $155.$