Let $N=abcd,$ meaning $N = 1000a+100b+10c+d.$

Now, given that $100b+10c+d = N/9$

So, $1000a = 8N/9 \implies 1125a = N.$

So, possible values of $a = \{1,2,3,4,5,6,7,8\}.$

(When $a=9, N=10000$ and this is no longer a 4 digit number)

Thus, no. of possible values of $N = 8.$