There are $n \ge 2$ numbers on the blackboard: $1, 2,..., n$. It is permitted to erase two of those numbers $x,y$ and write $2x - y$ instead. Find all values of $n$ such that it is possible to leave number $0$ on the blackboard after $n - 1$ such procedures.
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 \( n = 3 \).