, 2011

If we take any four consecutive even numbers and divide them by 8, we get remainders 0, 2, 4, 6 in some order. Thus there is only one number of the form $8k + 4$ among them which is divisible by 4 but not by 8. Hence if we take four even consecutive numbers

$$1000a + 100b + 10c + 2, \quad 1000a + 100b + 10c + 4,$$
$$1000a + 100b + 10c + 6, \quad 1000a + 100b + 10c + 8,$$

there is exactly one among these four which is divisible by 4 but not by 8. Now we can divide the set of all 4-digit even numbers with non-zero digits into groups of 4 such consecutive even numbers with $a, b, c$ nonzero. And in each group, there is exactly one number which is divisible by 4 but not by 8. The number of such groups is precisely equal to $9 \times 9 \times 9 = 729$, since we can vary $a, b, c$ in the set $\{1, 2, 3, 4, 5, 6, 7, 8, 9\}$.