Question 235, Find all integers $n \geq 3$, such that for any $n$ positive real numbers $a_{1} 、 a_{2} 、 \ldots 、 a_{n}$, if they satisfy $\operatorname{Max}\left\{a_{1} 、 a_{2} 、 \ldots 、 a_{n}\right\} \leq n \cdot \operatorname{Min}\left\{a_{1} 、 a_{2} 、 \ldots 、 a_{n}\right\}$, then among them, three numbers can be selected that can form the lengths of the three sides of an acute triangle.
For any $\mathrm{n} \leq 12$, in the sequence $1, 1, \sqrt{2}, \sqrt{3}, \sqrt{5}, \sqrt{8}, \sqrt{13}, \sqrt{21}, \sqrt{34}, \sqrt{55}, \sqrt{89}, 12$, taking the first $n$ numbers, we have $\operatorname{Max}\left\{a_{1}, a_{2}, \ldots, a_{n}\right\} \leq n \cdot \operatorname{Min}\left\{a_{1}, a_{2}, \ldots, a_{n}\right\}$, but no three numbers can form the sides of an acute triangle. Therefore, the condition is satisfied for $\mathrm{n} \geq 13$.
We will use proof by contradiction to show that for $n \geq 13$, the condition is always satisfied.
We sort the $n$ positive real numbers in non-decreasing order as $a_{1} \leq a_{2} \leq \ldots \leq a_{n}$. If any three of these numbers cannot form the sides of an acute triangle, then we have $a_{1}^{2} \leq a_{2}^{2}$ and $a_{k}^{2} + a_{k+1}^{2} \leq a_{k+2}^{2}$, where $1 \leq k \leq n-2$. Thus, we know that $a_{n}^{2} \geq f_{n} \cdot a_{1}^{2}$, which means $a_{n} \geq \sqrt{f_{n}} \cdot a_{1}$, where $f_{n}$ is the $n$-th term of the Fibonacci sequence. Furthermore, when $n \geq 13$, it is easy to show by mathematical induction that $\sqrt{f_{n}} > n$. Therefore, we have $a_{n} \geq \sqrt{f_{n}} \cdot a_{1} > n a_{1}$, which contradicts $\operatorname{Max}\left\{a_{1}, a_{2}, \ldots, a_{n}\right\} \leq n \cdot \operatorname{Min}\left\{a_{1}, a_{2}, \ldots, a_{n}\right\}$. Hence, for $n \geq 13$, the condition is always satisfied.
In conclusion, the values of $\mathrm{n}$ that satisfy the condition are all natural numbers not less than 13.