9. (20 points) For which $n>1$ do there exist such distinct natural numbers $a_{1}, a_{2}, \ldots, a_{n}$ that

$$
\operatorname{LCM}\left(a_{1}, a_{2}, \ldots, a_{n}\right)=a_{1}+a_{2}+\ldots+a_{n}
$$
Solution: Let there exist such different natural numbers $a$ and $b$ that $\operatorname{HOK}(a, b)=a+b$. Since

$$
\operatorname{HOK}(a, b) \vdots a \quad \text { and } \quad \operatorname{HOK}(a, b) \vdots b
$$

then

$$
a+b \vdots a \quad \text { and } \quad a+b \vdots b
$$

Therefore, $a \vdots b$ and $b \vdots a$, but this cannot be true because $a \neq b$. Now suppose that $n>2$ and consider the numbers

$$
a_{1}=1, a_{2}=2, a_{3}=2 \cdot 3, \ldots, a_{i}=2 \cdot 3^{i-2}, \ldots, a_{n-1}=2 \cdot 3^{n-3}, a_{n}=3^{n-2}
$$

All these numbers are distinct and their least common multiple is $2 \cdot 3^{n-2}$. Now let's find their sum

$$
\begin{gathered}
1+2+2 \cdot 3+\ldots+2 \cdot 3^{n-3}+3^{n-2}=1+2 \cdot\left(1+3+\ldots+3^{n-3}\right)+3^{n-2}= \\
=1+2 \cdot \frac{3^{n-2}-1}{3-1}+3^{n-2}=2 \cdot 3^{n-2}
\end{gathered}
$$

Answer: For all natural $n>2$.