How many distinct values of floor(N/i) exists for i=1 to N.

If $k \leqslant \sqrt{N}$, then $\lfloor N/\lfloor N/k\rfloor\rfloor = k$, since, letting $m = \lfloor N/k\rfloor$, we have $$mk\leqslant N < (m+1)k = mk + k \leqslant mk + m = m(k+1),$$ so that gives you $\lfloor \sqrt{N}\rfloor$ values. And the values of $\lfloor N/k\rfloor$ for $k \leqslant \lfloor \sqrt{N}\rfloor$ are all different, since

$$\frac{N}{k-1} - \frac{N}{k} = \frac{N}{k(k-1)} > 1$$

for $1 < k \leqslant \lfloor \sqrt{N}\rfloor$.

So you have either $2\lfloor \sqrt{N}\rfloor$ or $2\lfloor\sqrt{N}\rfloor - 1$ distinct values, depending on whether

$$N \geqslant \lfloor \sqrt{N}\rfloor(\lfloor\sqrt{N}\rfloor+1)$$

or not.