6. We will call a divisor $d$ of a natural number $n>1$ good if $d+1$ is also a divisor of $n$. Find all natural $n$ for which at least half of the divisors are good.
Answer: $2,6,12$

Solution. Consider such an $n$. If $d$ is a good divisor of $n$, then $d+1$ is also a divisor of $n$. However, the numbers $d$ and $d+1$ are coprime, so $n$ is divisible by $d(d+1)$ and, in particular, $n \geqslant d(d+1)>d^{2}$. Therefore, $dk^{3}$ and, thus, $k \leqslant 4$. It remains to check the numbers $n=\operatorname{HOK}(2,3)=6$ and $n=\operatorname{HOK}(2,3,4)=12$. Both of them work.