We prove that no positive divisor $d$ of $n$ that is greater or equal to $\sqrt{n}$ can be special: if $d$ is special, then $d + 1$ is also a divisor, so $n/d$ and $n/(d + 1)$ are both integers, which means that their difference is at least 1. Thus
$$\frac{n}{d} \geq \frac{n}{d+1} + 1,$$
which is equivalent to $n \ge d(d+1)$. But since $d(d+1) > d^2 \ge n$, this is a contradiction. Thus only divisors less than $\sqrt{n}$ can be special. Since divisors come in pairs ($a$ and $n/a$) such that one of them is less than $\sqrt{n}$ and one greater than $\sqrt{n}$ (when $n$ is a square, $\sqrt{n}$ is paired with itself), this means that at most half the divisors can be special.

If precisely half the divisors are special, then $n$ cannot be a square, and every divisor less than $\sqrt{n}$ has to be special. Thus 1 has to be a special divisor, meaning that 2 is a divisor (and thus also special), so 3 is a divisor, and so on, up to the greatest integer $k$ that is less than $\sqrt{n}$. Finally, $k$ is special, so $k + 1$ has to be a divisor as well. Since $k$ is the greatest divisor less than $\sqrt{n}$ and $k + 1$ the least divisor greater than $\sqrt{n}$, their product must be $n$, so $n = k(k + 1) = k^2 + k$. Moreover, $k - 1$ is also a divisor of $n = k^2 + k$ (unless $k = 1$), so it also divides $n - (k - 1)(k + 2) = k^2 + k - (k^2 + k - 2) = 2$.

This leaves us with $k = 1$, $k = 2$ and $k = 3$ as the only possibilities, giving us $n = 2$, $n = 6$ or $n = 12$. In all these cases, exactly half the divisors are special.