jmc

Note that this is the same as finding how many numbers with up to three digits do not contain 1. Since there are 10 total possible digits, and only one of them is not allowed (1), each place value has its choice of 9 digits, for a total of $9\ast 9\ast 9=729$ such numbers. However, we overcounted by one; 0 is not between 1 and 1000, so there are $\boxed{728}$ numbers. Note: As stated above, 020 is equivalent to 20, but 000 is not between 1 and 1000.