Solution: The integers satisfying the problem's condition are all primes.

First we check that all primes satisfy the condition of the problem. That is, if $p$ is a prime, then for all $n$, $1 \le n \le \frac{p}{2}$, $n$ divides the binomial coefficient $\binom{n}{p-2n}$. When $p=2$ this holds. For an odd prime $p$, take $n \in [1, \frac{p}{2}]$ and consider
$$(p-2n) \cdot \binom{n}{p-2n} = n \cdot \binom{n-1}{p-2n-1}.$$
Since $p \ge 2n$ and $p$ is odd, every factor in the above expression is nonzero. If $d = \gcd(p-2n, n)$, then $d$ divides $p$, but $d \le n < p$, so $d=1$. This shows that $p-2n$ and $n$ are coprime. Therefore $n$ divides the binomial coefficient $\binom{n}{p-2n}$.

Next, we prove that no composite number $m$ satisfies the property of the problem. Consider two cases:

(1) If $m = 2k, k > 1$, take $n = k$. Then $\frac{m}{3} \le n \le \frac{m}{2}$, but $\binom{k}{0} = 1$ cannot be divided by $k$.

(2) If $m$ is odd, then there exists an odd prime $p$ and an integer $k \ge 1$ such that $m = p(2k+1)$. Take $n = pk$, then from $k \ge 1$ we get $\frac{m}{3} \le n \le \frac{m}{2}$. However
$$\frac{1}{n} \binom{n}{m-2n} = \frac{1}{pk} \binom{pk}{p} = \frac{(pk-1)(pk-2)\cdots(pk-(p-1))}{p!}$$
is not an integer, because $p$ divides the denominator but does not divide the numerator.