Let $n \geq 3$ be an integer. Consider the set $A=\{1,2,3,\ldots,n\}$, in each move, we replace the numbers $i, j$ by the numbers $i+j$ and $|i-j|$. After doing such moves all of the numbers are equal to $k$. Find all possible values for $k$.
1. **Initial Setup and Problem Understanding**:
   We are given a set \( A = \{1, 2, 3, \ldots, n\} \) for \( n \geq 3 \). In each move, we replace two numbers \( i \) and \( j \) by \( i+j \) and \( |i-j| \). After a series of such moves, all numbers in the set become equal to \( k \). We need to find all possible values for \( k \).

2. **Prime Divisibility Argument**:
   Suppose \( k \) is the final number that all elements in the set become equal to. Let \( p \) be an odd prime number that divides \( k \). We will use a backward reasoning approach to understand the implications of this.

3. **Backward Reasoning**:
   If \( k \) is the final number, then \( p \) divides \( k \). Consider the reverse of the moves. If we reverse the steps, we would have started with numbers \( i \) and \( j \) such that \( p \) divides both \( i+j \) and \( |i-j| \).

4. **Divisibility by \( p \)**:
   Since \( p \) divides both \( i+j \) and \( |i-j| \), it follows that \( p \) must divide \( 2i \) and \( 2j \). This implies that \( p \) divides \( i \) and \( j \) (since \( p \) is odd and does not divide 2).

5. **Inductive Argument**:
   By induction, if \( p \) divides the final number \( k \), then \( p \) must divide all the numbers in the original set \( A \). However, this is a contradiction because the original set \( A \) contains numbers like 1, 2, 3, ..., which are not all divisible by any single odd prime \( p \).

6. **Conclusion**:
   Since no odd prime \( p \) can divide \( k \), the only possible value for \( k \) is 1. This is because 1 is the only number that is not divisible by any odd prime and can be the result of the operations described.

The final answer is \( \boxed{1} \).