2020 AMC 10A Problem 22

Write $1000=qn+r$, where $0\le r<n$. Then $\left\lfloor\dfrac{1000}{n}\right\rfloor=q$. The other two floors are usually also $q$, except that subtracting $1$ or $2$ from $1000$ crosses a multiple of $n$ when $r$ is small. For $n>1$, the sum is not divisible by $3$ exactly when $r=0$ or $r=1$. The case $r=0$ gives divisors of $1000$, excluding $1$, for $15$ values. The case $r=1$ gives divisors of $999$, excluding $1$, for $(3+1)(1+1)-1=7$ values. The total is $22$. Thus, A is the correct answer.