1. Base Cases:
- For $n = 1$, the only number on the blackboard is 1. It is impossible to perform any operation to get 0.
- For $n = 2$, the numbers are 1 and 2. We can perform the operation $(1, 2) \to (2 \cdot 1 - 2) = 0$.
- For $n = 3$, the numbers are 1, 2, and 3. We can try different operations, but it is impossible to get 0. For example:
$$(1, 2, 3) \to (0, 3) \to (0, 3) \to (0, 0) \to 0$$
This sequence does not work, and trying other combinations will also fail.

2. Inductive Step:
- Assume that for some $n \geq 2$, it is possible to leave the number 0 on the blackboard after $n - 1$ operations.
- We need to show that if it is possible for $n$, then it is also possible for $n + 2$.
- Consider the numbers $1, 2, \ldots, n, n+1, n+2$. Perform the operation on $n+1$ and $n+2$:
$$(n+1, n+2) \to (2(n+1) - (n+2)) = n$$
- Now we have the numbers $1, 2, \ldots, n, n$. Since we assumed it is possible to leave 0 with $n$ numbers, we can perform the operations on the first $n$ numbers to eventually get 0. The extra $n$ does not affect the process.

3. Conclusion:
- By induction, if it is possible for $n$, it is also possible for $n + 2$.
- We have shown that it is possible for $n = 2$ and impossible for $n = 1$ and $n = 3$.
- Therefore, it is possible for all $n \geq 2$ except $n = 3$.

The final answer is $\boxed{ n \geq 2 }$ except $n = 3$.