We are tasked with determining all integers $m \ge 2$ such that for every integer $n$ satisfying $\frac{m}{3} \leq n \leq \frac{m}{2}$, the binomial coefficient $\binom{n}{m-2n}$ is divisible by $n$.

To approach this problem, let's first consider the conditions on $n$. For a given $m$, the range for $n$ is $\left[\frac{m}{3}, \frac{m}{2}\right]$. Let $m - 2n = k$, so we need to ensure that the binomial coefficient $\binom{n}{k}$ is divisible by $n$.

### Understanding the Binomial Coefficient

The binomial coefficient $\binom{n}{k}$ can be expressed as:
$$\binom{n}{k} = \frac{n(n-1)(n-2)\cdots(n-k+1)}{k!}$$

For this to be divisible by $n$, the numerator must be divisible by $n$, which implies that $n$ divides at least one of the terms in the product $n(n-1)(n-2)\cdots(n-k+1)$.

### Analyzing when $\binom{n}{k}$ is Divisible by $n$

For the divisibility condition to be true for every $n$ in the specified range, one key requirement is to examine when $n$ appears as a factor in $\binom{n}{k}$. It often occurs that this condition is satisfied when $n$ is a prime number because in such cases, the factorial division in the binomial coefficient won't introduce a common factor across the range of $\left[\frac{m}{3}, \frac{m}{2}\right]$.

Therefore, if $m$ itself is structured such that every $n$ in the range can be non-composite, particularly being a prime, it inherently satisfies the condition that $n \mid \binom{n}{m-2n}$.

### Conclusion

Given this analysis, we can conclude that $m$ must be such that every applicable $n$ is inherently prime or acts divisibly in the factorial representation—specifically when $m$ is a prime number, this condition can be satisfied efficiently.

Thus, the required set of integers $m$ that ensures the condition is fulfilled for every $n$ in the specified range are all prime numbers. Therefore, the answer is:
$$\boxed{\text{all prime numbers}}$$