LCM of 3, 5, 6, and 9 = $90$

The smallest 4-digit number that is a multiple of 90 is 1080.

Given that the remainders when N is divided by 3, 5, 6, and 9 are 1, 3, 4, and 7 respectively, the differences between each divisor and remainder are the same: $(3-1) = (5-3) = (6-4) = (9-7) = 2$

Hence, the required number N is 2 less than the nearest multiple of 90. Therefore, N = $1080 - 2 = 1078$.