We need to prove that in a finite number of moves, all numbers on the board can be made equal using the given operations. Let's break down the solution step by step.

1. Base Case:
If $n = 1$, then there is only one number on the board, and it is trivially equal to itself. Therefore, the statement holds for $n = 1$.

2. General Case:
Assume $n \geq 2$. We will consider two main cases based on the greatest common divisor (gcd) of the numbers.

3. **Case 1: $\gcd(a, b) = 1$:**
- Take any two numbers $a$ and $b$ such that $\gcd(a, b) = 1$.
- Perform the operation on $(a, b)$ to obtain $\gcd(a, b) = 1$ and $\text{lcm}(a, b) - \gcd(a, b)$.
- Since $\gcd(a, b) = 1$, the new numbers are $1$ and $\text{lcm}(a, b) - 1$.
- Now, we can repeatedly apply the operation on any number $x$ and $1$:
$$\gcd(x, 1) = 1 \quad \text{and} \quad \text{lcm}(x, 1) - \gcd(x, 1) = x - 1$$
- This decreases $x$ by 1 in each step until $x$ becomes 1. Thus, all numbers can be reduced to 1.

4. **Case 2: $\gcd(a, b) \neq 1$ for all pairs $(a, b)$:**
- Assume there is no pair $(a, b)$ such that $\gcd(a, b) = 1$.
- If there exists a number $c$ such that $\gcd(a, c) \perp \gcd(b, c)$ (i.e., $\gcd(\gcd(a, c), \gcd(b, c)) = 1$), we can apply the operation on $(a, c)$ and then on $(\gcd(a, c), b)$ to eventually get 1 on the board. This reduces to Case 1.
- If no such $c$ exists, then $\gcd(\{a_i\}_n) \neq 1$ for all $i$. Let $d = \gcd(\{a_i\}_n)$.
- Take any two numbers $a$ and $b$ such that $\gcd(a, b) = d$. Perform the operation on $(a, b)$ to obtain $d$ and $\text{lcm}(a, b) - d$.
- Now, take any number $x$ and note that:
$$\gcd(d, x) = d \quad \text{and} \quad \text{lcm}(x, d) - \gcd(x, d) = x - d$$
- This decreases $x$ by $d$ in each step until $x$ becomes $d$. Repeating this process for all numbers, we can make all numbers equal to $d$.

Thus, in both cases, we can make all numbers on the board equal in a finite number of moves.

$\blacksquare$