Restricted divisiblity

Let us first make an observation.

If we take any $4$ consecutive even numbers(say $8k,8k+2,8k+4,8k+6$ )then clearly only one among them will be divisible by 4 and not by $8$ ,ie,of the form $8k+4$ .

Since here we consider only the numbers having $4$ digits,let the numbers be

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

So it is quite intuitive that we can divide the set of even numbers into subsets of cardinality $4$ such that each set contains only one number which is divisible by $4$ but not by $8$ .Hence there is a bijection between the number of permissible integers and the number of such $4$ element subsets.Now for each choice of $a,b,c,d$ we get a new $4$ element permissible subset of even numbers and hence the number of possible subsets is equal to the number of such 4 digit integers.

Since $a,b,c,d \epsilon \{1,2,3,4,5,6,7,8,9\}$ ,there are 9 choices for each of $a,b,c,d$ .Hence the total number of such integers is $9*9*9 = 729$ .

Hence our answer is $\boxed{729}$ .

$\textbf{Note:}$ The cardinality of a set denotes the number of elements in the set.