, 2005

Solution:

The answer is 1002. If the units digit of an integer $n$ is even, then the units digit of $n+1$ is odd, while all the other digits of $n$ and $n+1$ are the same. Therefore, if $n$ has an even number of even digits, then $n+1$ has an odd number of them; and if $n$ has an odd number of even digits, then $n+1$ has an even number of them. Hence, exactly one of $n$ and $n+1$ has an odd number of even digits. Dividing the integers between 2 and 2005 into the 1002 pairs $(2,3),(4,5),(6,7), \ldots,(2004,2005)$, we deduce that exactly 1002 of them have an odd number of even digits. Since 1 has no even digits, and therefore is not to be counted, the answer is 1002.