Clearly, $n=1$ fails. Except for the special case of $n=1$,
$$\left\lfloor \frac{1000}{n} \right\rfloor - \left\lfloor \frac{998}{n} \right\rfloor$$
equals either $0$ or $1$. If it equals $0$, this implies that $\left\lfloor \frac{998}{n} \right\rfloor = \left\lfloor \frac{999}{n} \right\rfloor = \left\lfloor \frac{1000}{n} \right\rfloor$, 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 $3 \left\lfloor \frac{999}{n} \right\rfloor \pm 1$, so it is never a multiple of $3$. Thus, we are looking for all $n \neq 1$ such that
$$\left\lfloor \frac{1000}{n} \right\rfloor - \left\lfloor \frac{998}{n} \right\rfloor = 1.$$
This implies that either
$$\left\lfloor \frac{998}{n} \right\rfloor + 1 = \left\lfloor \frac{999}{n} \right\rfloor,$$
or
$$\left\lfloor \frac{999}{n} \right\rfloor + 1 = \left\lfloor \frac{1000}{n} \right\rfloor.$$
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
$$\frac{998}{n} < a \leq \frac{999}{n} \implies 998 < an \leq 999 \implies an = 999 \implies a = \frac{999}{n} \implies n | 999.*$$
Analogously, the second equation implies that
$$n | 1000.$$
So our only $n$ that satisfy this condition are $n \neq 1$ 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 $23-1 = \boxed{\textbf{(A) } 22}.$

*While this observation may seem strange, it is actually "trivial by intuition" to go straight from $\left\lfloor \frac{998}{n} \right\rfloor + 1 = \left\lfloor \frac{999}{n} \right\rfloor$ to $n | 999$. In fact, "trivial by intuition" is basically a good summary of the solution to this entire problem.

~ihatemath123