Find all integers $n \geq 3$ such that among any $n$ positive real numbers $a_{1}, a_{2}, \ldots, a_{n}$ with
$$
\max \left(a_{1}, a_{2}, \ldots, a_{n}\right) \leq n \cdot \min \left(a_{1}, a_{2}, \ldots, a_{n}\right),
$$
there exist three that are the side lengths of an acute triangle.
The answer is all $n \geq 13$.
Define $\left(F_{n}\right)$ as the sequence of Fibonacci numbers, by $F_{1}=F_{2}=1$ and $F_{n+1}=$ $F_{n}+F_{n-1}$. We will find that Fibonacci numbers show up naturally when we work through the main proof, so we will isolate the following calculation now to make the subsequent solution easier to read.
Claim - For positive integers $m$, we have $F_{m} \leq m^{2}$ if and only if $m \leq 12$.

Proof. A table of the first 14 Fibonacci numbers is given below.
\begin{tabular}{rrrrrrrrrrrrrr}
$F_{1}$ & $F_{2}$ & $F_{3}$ & $F_{4}$ & $F_{5}$ & $F_{6}$ & $F_{7}$ & $F_{8}$ & $F_{9}$ & $F_{10}$ & $F_{11}$ & $F_{12}$ & $F_{13}$ & $F_{14}$ \\
\hline 1 & 1 & 2 & 3 & 5 & 8 & 13 & 21 & 34 & 55 & 89 & 144 & 233 & 377
\end{tabular}

By examining the table, we see that $F_{m} \leq m^{2}$ is true for $m=1,2, \ldots 12$, and in fact $F_{12}=12^{2}=144$. However, $F_{m}>m^{2}$ for $m=13$ and $m=14$.

Now it remains to prove that $F_{m}>m^{2}$ for $m \geq 15$. The proof is by induction with base cases $m=13$ and $m=14$ being checked already. For the inductive step, if $m \geq 15$ then we have
$$
\begin{aligned}
F_{m} & =F_{m-1}+F_{m-2}>(m-1)^{2}+(m-2)^{2} \\
& =2 m^{2}-6 m+5=m^{2}+(m-1)(m-5)>m^{2}
\end{aligned}
$$
as desired.
We now proceed to the main problem. The hypothesis $\max \left(a_{1}, a_{2}, \ldots, a_{n}\right) \leq n$. $\min \left(a_{1}, a_{2}, \ldots, a_{n}\right)$ will be denoted by ( $\dagger$ ).

Proof that all $n \geq 13$ have the property. We first show now that every $n \geq 13$ has the desired property. Suppose for contradiction that no three numbers are the sides of an acute triangle. Assume without loss of generality (by sorting the numbers) that $a_{1} \leq a_{2} \leq \cdots \leq a_{n}$. Then since $a_{i-1}, a_{i}, a_{i+1}$ are not the sides of an acute triangle for each $i \geq 2$, we have that $a_{i+1}^{2} \geq a_{i}^{2}+a_{i-1}^{2}$; writing this out gives
$$
\begin{array}{l}
a_{3}^{2} \geq a_{2}^{2}+a_{1}^{2} \geq 2 a_{1}^{2} \\
a_{4}^{2} \geq a_{3}^{2}+a_{2}^{2} \geq 2 a_{1}^{2}+a_{1}^{2}=3 a_{1}^{2} \\
a_{5}^{2} \geq a_{4}^{2}+a_{3}^{2} \geq 3 a_{1}^{2}+2 a_{1}^{2}=5 a_{1}^{2} \\
a_{6}^{2} \geq a_{5}^{2}+a_{4}^{2} \geq 5 a_{1}^{2}+3 a_{1}^{2}=8 a_{1}^{2}
\end{array}
$$
and so on. The Fibonacci numbers appear naturally and by induction, we conclude that $a_{i}^{2} \geq F_{i} a_{1}^{2}$. In particular, $a_{n}^{2} \geq F_{n} a_{1}^{2}$.

However, we know $\max \left(a_{1}, \ldots, a_{n}\right)=a_{n}$ and $\min \left(a_{1}, \ldots, a_{n}\right)=a_{1}$, so ( $\dagger$ ) reads $a_{n} \leq n \cdot a_{1}$. Therefore we have $F_{n} \leq n^{2}$, and so $n \leq 12$, contradiction!

Proof that no $n \leq 12$ have the property. Assume that $n \leq 12$. The above calculation also suggests a way to pick the counterexample: we choose $a_{i}=\sqrt{F_{i}}$ for every $i$. Then $\min \left(a_{1}, \ldots, a_{n}\right)=a_{1}=1$ and $\max \left(a_{1}, \ldots, a_{n}\right)=\sqrt{F_{n}}$, so ( $\dagger$ ) is true as long as $n \leq 12$. And indeed no three numbers form the sides of an acute triangle: if $i<j<k$, then $a_{k}^{2}=F_{k}=F_{k-1}+F_{k-2} \geq F_{j}+F_{i}=a_{j}^{2}+a_{i}^{2}$.