Question 2. $n$ is a positive integer greater than 1, $a_{1}, a_{2}, \ldots, a_{n}$ are $n$ distinct positive integers. Let $M=\left\{\left(a_{i}, a_{j}\right),\left[a_{i}, a_{j}\right] \mid 1 \leq i<j \leq n\right\}$, find the minimum number of elements in $M$.

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Answer 2. For any finite set of positive integers $A$, we have $M_{A}=\left\{\left(a_{i}, a_{j}\right),\left[a_{i}, a_{j}\right] \mid a_{i} \neq a_{j} \in A\right\}$, which is a simplified notation for $M$ in the problem.

Fix a positive integer $a$ greater than 1, and take $a_{i}=a^{i}$, then $M=\left\{a_{1}, \ldots, a_{n}\right\}$, at this time the number of elements in $M$ is $n$. We will prove that the number of elements in $M$ cannot be less than $n$.

If $n=2$, if $|M|<2$. Then for any $a_{i}, a_{j} \in S$, we have $v_{p}\left(\left(a_{i}, a_{j}\right)\right)=v_{p}\left(\left[a_{i}, a_{j}\right]\right)=s$. And for any $a_{i}, a_{j} \in T$, we have $v_{p}\left(\left(a_{i}, a_{j}\right)\right), v_{p}\left(\left[a_{i}, a_{j}\right]\right)<s$. That is, $M_{S}$ and $M_{T}$ are disjoint. By mathematical induction, we know $\left|M_{S}\right| \geq|S|,\left|M_{T}\right| \geq|T|$, then $M \supset M_{S} \cup M_{T}$ implies $|M| \geq|S|+|T|=n$.
2. $|S|=1$. Suppose $S=\{a\}$. Then for any $a_{i}, a_{j} \in T$, we have $v_{p}\left(\left(a_{i}, a_{j}\right)\right), v_{p}\left(\left[a_{i}, a_{j}\right]\right)<s$. But $v_{p}\left(\left[a_{i}, a\right]\right)=s$, which means $\left[a_{i}, a\right] \notin M_{T}$, so $|M| \geq\left|M_{T}\right|+1$. Similarly, by induction $\left|M_{T}\right| \geq|T|$, then $|M| \geq\left|M_{T}\right|+1 \geq|T|+1=n$.