How Many Numbers?

Find the number of integers $N$ lying between $2$ and $100$ (inclusive) such that for all integers $m$ such that $\dfrac{N}{3} \leq m \leq \dfrac{N}{2},$ $\dbinom{m}{N-2m}$ is a multiple of $m.$

Details and assumptions

• $N=2$ is considered a solution, since the only integer $m$ such that $\dfrac{2}{3} \leq m \leq \dfrac{2}{2}$ is $1,$ and $\dbinom{2}{2 - 2 \cdot 1}$ is a multiple of $1.$

• This problem is not original.

@Calvin sir: The intended expression is $\dbinom{m}{N-2m}.$ This is valid since $m \geq N - 2m \geq 0$ from the condition $\dfrac{N}{3} \leq m \leq \dfrac{N}{2}.$ Roger Lu has commented why $\dbinom{N}{N-2m}$ doesn't work. Please correct me if I am wrong.