Find all such odd natural $n>1$, that for any coprime divisors $a$ and $b$ of $n$, the number $a+b-1$ is also a divisor of $n$.
Let $p$ be the smallest prime divisor of the number $n$. Represent $n$ in the form $p^{m} k$, where $k$ is not divisible by $p$. By the condition, the number $p+k-1$ is a divisor of $n$.

If $(p+k-1, k)>1$, then $(p-1, k)=(l, k)>1$. Thus, the number $k$ has some divisor $d, 2 \leq d \leq p-$

1. Contradiction with the choice of the number $p$. Therefore, $p+k-1=p^{\alpha}$.

Suppose $\alpha \geq 2$. Then $p^{2}$ and $k-$ are coprime divisors of the number $n$, so $p^{2}+k-1$ is a divisor of the number $n$. At the same time, $p^{2}+k-1$ is coprime with $k$, since otherwise $k$ has a common divisor with $p^{2}-1=2(p-1) \cdot 1 / 2(p+$ 1), which again contradicts the choice of the number $p$. Therefore, $p^{2}+k-1=p^{\beta}$, where $\beta \geq 3$. But then $p^{\beta}=p^{2}+k-1=p^{2}+(p$ $+k-1)-p=p\left(p+p^{\alpha-1}-1\right)$, which is not divisible by $p^{2}$. Contradiction.

Therefore, $k=1$, that is, $n=p^{m}$. It is not difficult to verify that all such numbers satisfy the condition.

## Answer

$n$ is a power of an odd prime number.