jmc

An integer is divisible by $8$ if and only if the number formed from its last three digits is divisible by $8$. Thus, the number of possibilities for the last three digits is equal to the number of three-digit multiples of $8$. Since $1000=8\cdot 125$, we find that there are $125$ such multiples. Since the thousands digit of our four-digit integer must be nonzero, there are $9$ possibilities for the thousands digit. Altogether, $9\cdot 125=\boxed{1125}$ four-digit integers are divisible by $8$.