The total number of 4-digit numbers whose greatest common divisor with 18 is 3 is ______.

We need to count 4-digit numbers $$n$$ (from 1000 to 9999) such that $$\gcd(n, 18) = 3$$. Since $$18 = 2 \cdot 3^2$$, the condition $$\gcd(n, 18) = 3$$ requires: (i) $$3 \mid n$$ (so $$\gcd$$ includes at least one factor of 3), (ii) $$9 \nmid n$$ (otherwise $$\gcd$$ would include $$9$$), and (iii) $$2 \nmid n$$ (otherwise $$\gcd$$ would include a factor of 2).

So we need $$n$$ to be odd, divisible by 3, but not divisible by 9.

The number of odd 4-digit numbers is $$\frac{9999 - 1001}{2} + 1 = 4500$$. The number of odd multiples of 3 in this range equals $$\frac{4500}{3} = 1500$$ (since among any three consecutive odd numbers, exactly one is divisible by 3).

The number of odd multiples of 9 in $$[1000, 9999]$$: The multiples of 9 in this range are $$1008, 1017, \ldots, 9999$$, totalling $$\frac{9999 - 1008}{9} + 1 = 1000$$. Exactly half of these are odd (since consecutive multiples of 9 alternate parity as 9 is odd), giving 500 odd multiples of 9.

Therefore, the count of 4-digit numbers with $$\gcd(n, 18) = 3$$ is $$1500 - 500 = 1000$$.