$$$\text{LCM}$$$ is quite uncomfortable. Rephrase the problem using $$$\text{GCD}$$$.

Use $$$\text{LCM}(a, b) = \frac{ab}{\text{GCD}(a, b)}$$$. Now you can solve the problem separately for each $$$k = \text{GCD}(a_i, a_j)$$$ ($$$i \neq j$$$). What are the optimal values of $$$a_i, a_j$$$?

For a fixed $$$k$$$, it's optimal to choose the minimum $$$a_i, a_j$$$ that are multiples of $$$k$$$. How to find those minimums? Does it matter if, actually, $$$\text{GCD}(a_i, a_j) > k$$$?

You have to minimize $$$\frac{a_ia_j}{\text{GCD}(a_i, a_j)}$$$ ($$$i \neq j$$$).

For each $$$k = \text{GCD}(a_i, a_j)$$$, $$$a_i, a_j$$$ have to be multiples of $$$k$$$. Note that, if $$$\text{GCD}(a_i, a_j) = h > k$$$, you end up calculating a wrong value of $$$\text{LCM}(a_i, a_j)$$$, but it doesn't matter because it's never optimal; you will calculate the correct (smaller) value of the $$$\text{LCM}$$$ while considering $$$\text{GCD}(a_i, a_j) = h$$$ .

So, for each $$$k$$$, it's enough to consider the smallest $$$a_i, a_j$$$ that are multiples of $$$k$$$. You can find them like in Solution 1 or Solution 3 (instead of the sum, you should keep the two minimum elements).

Once again, use $$$\text{GCD}$$$ instead of $$$\text{LCM}$$$ and solve the problem for each fixed $$$\text{GCD}$$$.

Let $$$\displaystyle g(k) = \sum_{k | a_i, k | a_j, i < j}a_ia_j$$$. Can you calculate it with Solution 1 (or Solution 3)?

Let $$$g_1(k) = \sum_{k | a_i}a_i$$$, $$$g_2(k) = \sum_{k | a_i}a_i^2$$$ (they can be calculated with Solution 1 or Solution 3). Then, $$$g(k) = \frac{1}{2}(g_1(k)^2 - g_2(k))$$$ (it's easy to prove).

Now the problem is that some products $$$a_ia_j$$$ considered in $$$g(k)$$$ may have $$$\text{GCD}(a_i, a_j) > k$$$. How to fix that?

You have to calculate the sum of $$$\frac{a_ia_j}{\text{GCD}(a_i, a_j)}$$$ ($$$i < j$$$).

For each $$$k = \text{GCD}(a_i, a_j)$$$, $$$a_i, a_j$$$ have to be multiples of $$$k$$$. Let's start by calculating $$$\displaystyle g(k) = \sum_{k | a_i, k | a_j, i < j}a_ia_j$$$, then we will remove pairs such that $$$\text{GCD}(a_i, a_j) > k$$$.

Let $$$g_1(k) = \sum_{k | a_i}a_i$$$, $$$g_2(k) = \sum_{k | a_i}a_i^2$$$ (they can be calculated with Solution 1 or Solution 3, using $$$b_i = a_i$$$ and $$$b_i = a_i^2$$$ respectively). Note that $$$\displaystyle g_2(k) = \sum_{k | a_i, k | a_j}a_ia_j$$$. Then, $$$g(k) = \frac{1}{2}(g_1(k)^2 - g_2(k))$$$.

Finally, let $$$\displaystyle f(k) = \sum_{\text{GCD}(a_i, a_j) = k, i < j}a_ia_j$$$. Let's calculate $$$f(k)$$$ for $$$k$$$ from $$$m$$$ to $$$1$$$. Then, $$$f(k) = g(k) - \sum_{i=2}^{\lfloor m/k \rfloor} f(ik)$$$ (it's the same idea as the Solution 1).

Suppose you want the last number of the blackboard to be $$$k$$$. Is there an optimal order of operations (such that, if $$$k$$$ can be reached, you can also reach it with that order of operations)?

It's optimal to use all the $$$\text{GCD}$$$ operations before all the $$$\text{MIN}$$$ operations. So, how does the last number of the blackboard look like?

The last number of the blackboard is $$$\leq \min(a_i)$$$, and it's the $$$\text{GCD}$$$ of a subset of $$$a$$$. How to count the possible final numbers?

If $$$k$$$ is the $$$\text{GCD}$$$ of some subset, it's also the $$$\text{GCD}$$$ of $$$S_k = \{a_i \mid k | a_i\}$$$. How to calculate the $$$\text{GCD}$$$ of $$$S_k$$$ for each $$$k$$$?

It's easy to prove that the final number on the blackboard is $$$\leq \min(a_i)$$$, and it's the $$$\text{GCD}$$$ of a subset of $$$a$$$.

For a fixed $$$k$$$, a necessary and sufficient condition for it to be the $$$\text{GCD}$$$ of a subset is to be the $$$\text{GCD}$$$ of the "worst-case" subset, namely $$$S_k = \{a_i \mid k | a_i\}$$$. Let $$$f(k)$$$ be such $$$\text{GCD}$$$.

Initially, all $$$f(k)$$$ are set to $$$0$$$. For each $$$a_i$$$, iterate over its divisors $$$j$$$ (with Solution 2) and set $$$f(j) := \text{GCD}(f(j), a_i)$$$

The answer is the number of $$$k \leq \min(a_i)$$$ such that $$$f(k) = k$$$.