Why three-digit integers only?

Let's rewrite $N$ without the "#" function:

$N=\frac{1000A+B}{B}=\frac{1000A}{B}+1$

To minimize $N$ , we must minimize $A$ and maximize $B$ such that $B$ divides $1000A$ . First, notice that if we let $A=102+3m$ and $B=850+25m$ , then:

$N = \frac{1000(102+3m)}{850+25m}+1= 1000\frac{3(m+34)}{25(m+34)}+1= 121$

If $A\geq125$ , then $N > \frac{125000}{1000}+1=126$ , so we only need to check values of $A<125$ .

I personally checked all of the values of $A<125$ , and if $A≠102+3m$ , then $N$ achieves its maximum when $B=8A$ , therefore $N=\frac{1000A}{8A}+1=126$ in these situations, therefore our minimum $N$ must be $\boxed{121}$ .

To be honest, I was lucky solving this one and I didn't try to get an approach. I started to decrease $B$ from $1000$ and increase $A$ from $100$ , for getting a pair $(A, B)$ such that $B|1000 A$ . First pair that came to my mind was $(A, B)=(108, 900)$ . It gave me the answer as $N=121$ and it was right.