1. Find the sum of all even natural numbers $n$ for which the number of divisors (including 1 and $n$ itself) is equal to $\frac{n}{2}$. (For example, the number 12 has 6 divisors: $1,2,3,4,6,12$.)
Solution. Let the canonical decomposition of the number $n$ be: $n=2^{t_{1}} \cdot 3^{t_{2}} \cdot 5^{t_{3}} \cdots \cdots \cdot p^{t_{k}}$. Then the number of divisors of the number $n$ is $\left(t_{1}+1\right)\left(t_{2}+1\right)\left(t_{3}+1\right) \cdots\left(t_{k}+1\right)$. From the condition of the problem, we have the equality

$$
\left(t_{1}+1\right)\left(t_{2}+1\right)\left(t_{3}+1\right) \cdots\left(t_{k}+1\right)=2^{t_{1}-1} \cdot 3^{t_{2}} \cdot 5^{t_{3}} \cdots \cdots \cdot p^{t_{k}} \cdot\left({ }^{*}\right)
$$

Note that $2^{t_{1}-1}>t_{1}+1$ for $t_{1} \geq 4, 3^{t_{2}}>t_{2}+1$ for $t_{2} \geq 1, \ldots, p^{t_{k}}>t_{k}+1$ for $t_{k} \geq 1$. Therefore, $t_{1}$ can take the values 1, 2, or 3. Substituting the specified values into the equality $(*)$, we find that $n=8$ or $n=12$.

Answer: 20.