Which are those positive integers $n$ that have $\frac{n}{2}$ positive divisors?
I. solution. The divisors of a positive number can be paired so that the product of each pair is the number itself. The smaller factor of the product is at most $\sqrt{n}$. (Indeed, if both factors were greater than $\sqrt{n}$, their product would be greater than $n$. For example, the divisors of 12 can be paired as follows: 1, $12 ; 2,6 ; 3,4$, and $3<\sqrt{12}$, or the divisors of 36 are 1, $36 ; 2,18,3,12,4,9 ;$ and the unpaired $6=\sqrt{36}$.)

Thus, the number of divisors of $n$ can be at most $2[\sqrt{n}]$. (The square brackets denote the integer part of the number - the largest integer not greater than the number.)

The positive integers we are looking for can therefore only be among those values of $n$ for which

$$
\frac{n}{2} \leq 2[\sqrt{n}], \quad \text { so } \quad n \leq 4[\sqrt{n}] \leq 4 \sqrt{n}
$$

From this, $\sqrt{n} \leq 4$, so the value of $n$ can be at most 16, and since in our case $\frac{n}{2}$ is a positive integer, the value of $n$ can be $2,4,6,8,10,14$ or 16.

If we list the divisors of these, we can see that only 8 and 12 satisfy the requirements of the problem.

II. solution. Since the number of divisors is even, $n=2k$ in form. The number of divisors of 2 is $2 \neq \frac{2}{2}$ and the number of divisors of 4 is $3 \neq \frac{4}{2}$, so $k \geq 3$. According to this, $1 \leq k-2<k-1<k$. If neither $k-1$ nor $k-2$ were a divisor of $(2k)$, then $(2k)$ would have at most $k-2$ divisors among the numbers from 1 to $k$. Considering that the double of $(k+1)$ is already greater than $(2k)$, the only divisor of $n$ among the numbers greater than $k$ would be $2k$; and thus the number of divisors would be at most $k-1$. The number of divisors can therefore only be at least $k$ if one of these two numbers is a divisor of $n$.

If $2k$ is divisible by $(k-1)$, then $2=2k-2(k-1)$ is also divisible by $(k-1)$. If $2k$ is divisible by $(k-2)$, then $4=2k-2(k-2)$ is also divisible by $(k-2)$. Since $k \geq 3$, in the first case $k-1=2$; in the second case $k-2$ is either 1, 2, or 4 - since these are all the positive divisors of 4. Accordingly, the possible values for $k$ are $3$; and 3, 4, and 6. Therefore, the possible solutions are 6, 8, and 12. As we saw in the previous solution, only $n=8$ and $n=12$ meet the requirements.