Let us divide!

Let required set of numbers be N={n}

Condition 1 - N should be divisible by 3 and 7, which means it should be divisible by 21

Condition 2 - N should not be divisible by 5 and 9.

We have to identify and take out the common numbers with those in condition 1. As 3 has already been counted once in condition 1, so this is equivalent to saying that N should be divisible by 21 but not by 21 x 5 x 3 = 315

Hence the required count is [ $\frac{1000}{21}$ ] - [ $\frac{1000}{315}$ ] = 47 - 3 = 44