9. Let $n \geq 2$ be a natural number. For a subset $F$ with $n$ elements of $\{1, \ldots, 2 n\}$, we define $m(F)$ as the minimum of all $\operatorname{kgV}(x, y)$, where $x$ and $y$ are two distinct elements of $F$. Find the maximum value that $m(F)$ can achieve.
Solution: We first show that $m(F)$ reaches its maximum for $F=\{n+1, \ldots, 2 n\}$. Indeed, let $G$ be another subset and $x \leq n$ an element of $G$.

(a) If $2 x$ is also an element of $G$, then $m(G) \leq \operatorname{kgV}(x, 2 x)=2 x$. On the other hand, since every element of $F$ is strictly greater than $n$, none divides another, and thus for any pair $x \neq y$ of elements of $F, \operatorname{kgV}(x, y)>2 n$, so $m(F)>m(G)$.

(b) If $2 x$ is not an element of $G$, for any $y \neq x$ element of $G$, we have $\operatorname{kgV}(x, y) \leq \operatorname{kgV}(2 x, y)$, so if $G^{\prime}$ is the set obtained from $G$ by replacing $x$ with $2 x$, we then have that $m(G) \leq m\left(G^{\prime}\right)$.

The maximum is therefore reached when $G$ has no element among $\{1, \ldots, n\}$, i.e., when $G=F$ (even though it can also be reached for other sets $G$).

It is therefore sufficient to calculate $m(F)$ for $F=\{n+1, \ldots, 2 n\}$. Let $x_{0}<y_{0}$ be elements of $F$ such that $m(F)=\operatorname{kgV}\left(x_{0}, y_{0}\right)$. If $d=\operatorname{ggT}\left(x_{0}, y_{0}\right)$, we then have that $m(F)=\frac{x_{0} \cdot y_{0}}{d}$. Now, since $y_{0}=x_{0}+k \cdot d$ for some $k \in \mathbb{N}$ and since $\left(x_{0}, y_{0}\right)$ minimize $\operatorname{kgV}\left(x_{0}, y_{0}\right)=\frac{x_{0} \cdot y_{0}}{d}$, we must have $k=1$, so $m(F)=x_{0} \cdot\left(\frac{x_{0}}{d}+1\right)$.

We start by calculating that for $n \leq 4, m(F)$ is successively 12, 12, and 24. Now for $n \geq 5$, we treat $n$ even and $n$ odd separately:

- If $n$ is odd, for $x_{0}=n+1$ and $d=\frac{n+1}{2}$, we get $y_{0}=\frac{3(n+1)}{2}$ and since $x_{0}$ is the smallest element of $F$ and for any choice of $x_{0}$ in $F \frac{x_{0}}{d} \geq 2$, we have that $m(F)=\operatorname{kgV}\left(x_{0}, y_{0}\right)=3(n+1)$. Moreover, $y_{0}=\frac{3(n+1)}{2} \leq 2 n$ because this inequality is equivalent to $n \geq 3$, which is true since we consider $n \geq 5$.
- If $n$ is even, for $x_{0}=n+2$ and $d=\frac{n+2}{2}$, we get $y_{0}=\frac{3(n+2)}{2}$. Since $\frac{x_{0}}{d}=2$, the only other candidate for $x_{0}$ would be $n+1$ but since $n+1$ is odd, we would then have $\frac{x_{0}}{d^{\prime}} \geq 3$, then $(n+1) \cdot\left(\frac{n+1}{d^{\prime}}+1\right) \geq(n+1) \cdot 4 \geq(n+2) \cdot 3=(n+2) \cdot\left(\frac{n+2}{d}+1\right)$ because this inequality is equivalent to $n \geq 2$. We therefore have that $m(F)=\operatorname{kgV}\left(x_{0}, y_{0}\right)=3(n+2)$. Finally, we observe that $y_{0}=\frac{3(n+2)}{2} \leq 2 n$ because the inequality is equivalent to $n \geq 6$.

The solution is therefore $3(n+1)$ if $n$ is odd and $3(n+2)$ if $n$ is even, with the only exception being $n=4$ where the solution is 24.