imc

Clearly, $n=1$ fails. Except for the special case of $n=1$, $$\lfloor \frac{1000}{n}\rfloor -\lfloor \frac{998}{n}\rfloor$$ equals either $0$ or $1$. If it equals $0$, this implies that $\lfloor \frac{998}{n}\rfloor =\lfloor \frac{999}{n}\rfloor =\lfloor \frac{1000}{n}\rfloor$, so their sum is clearly a multiple of $3$, so this will always fail. If it equals $1$, the sum of the three floor terms is $3\lfloor \frac{999}{n}\rfloor \pm 1$, so it is never a multiple of $3$. Thus, we are looking for all $n\ne 1$ such that $$\lfloor \frac{1000}{n}\rfloor -\lfloor \frac{998}{n}\rfloor =1.$$ This implies that either $$\lfloor \frac{998}{n}\rfloor +1=\lfloor \frac{999}{n}\rfloor ,$$ or $$\lfloor \frac{999}{n}\rfloor +1=\lfloor \frac{1000}{n}\rfloor .$$ Let's analyze the first equation of these two. This equation is equivalent to the statement that there is a positive integer $a$ such that $$\frac{998}{n}<a\le \frac{999}{n}⟹998<an\le 999⟹an=999⟹a=\frac{999}{n}⟹n|999.\ast$$ Analogously, the second equation implies that $$n|1000.$$ So our only $n$ that satisfy this condition are $n\ne 1$ that divide $999$ or $1000$. Using the method to find the number of divisors of a number, we see that $999$ has $8$ divisors and $1000$ has $16$ divisors. Their only common factor is $1$, so there are $8+16-1=23$ positive integers that divide either $999$ or $1000$. Since the integer $1$ is a special case and does not count, we must subtract this from our $23$, so our final answer is $23-1=\boxed{22}.$