Given integer $n \geq 3$. $1, 2, \ldots, n$ were written on the blackboard. In each move, one could choose two numbers $x, y$, erase them, and write down $x + y, |x-y|$ in the place of $x, y$. Find all integers $X$ such that one could turn all numbers into $X$ within a finite number of moves.
1. **Initial Observations**:
   - We start with the numbers \(1, 2, \ldots, n\) on the blackboard.
   - In each move, we can choose two numbers \(x\) and \(y\), erase them, and write down \(x + y\) and \(|x - y|\).
   - We need to determine all integers \(X\) such that all numbers can be turned into \(X\) within a finite number of moves.

2. **Necessary Conditions**:
   - Consider the largest number on the blackboard at any point in time. This number either increases or stays constant.
   - Initially, the largest number is \(n\). Therefore, the final number \(X\) must be at least \(n\), i.e., \(X \geq n\).

3. **Divisibility by Odd Primes**:
   - Suppose \(X\) is divisible by an odd prime \(p\). Consider reversing the operations on the final tuple \((X, X, \ldots, X)\).
   - The reverse operation is \((a, b) \mapsto \left(\frac{a+b}{2}, \frac{|a-b|}{2}\right)\).
   - If \(p\) divides \(X\), then \(p\) must divide all numbers on the blackboard at all times. This is impossible since \(1\) was on the board initially.
   - Therefore, \(X\) cannot be divisible by any odd prime, implying \(X\) must be a power of 2.

4. **Sufficient Conditions**:
   - Let \(\ell\) be the smallest positive integer such that \(2^\ell \geq n\).
   - We need to show that \((2^m, 2^m, \ldots, 2^m)\) is attainable whenever \(m \geq \ell\).

5. **Doubling Technique**:
   - We can perform the operation \((x, y) \mapsto (x+y, |x-y|) \mapsto (2x, 2y)\).
   - Specifically, \((0, x) \mapsto (0, 2x)\), meaning that as long as there is a zero in the tuple, any arbitrary number can be doubled.

6. **Inductive Step**:
   - Suppose we can reach \((2^x, 2^x, \ldots, 2^x)\). Then:
     \[
     (2^x, 2^x, \ldots, 2^x) \mapsto (0, 2^{x+1}, 2^x, 2^x, \ldots, 2^x) \mapsto (0, 2^{x+1}, 2^{x+1}, \ldots, 2^{x+1}) \mapsto (2^{x+1}, 2^{x+1}, \ldots, 2^{x+1})
     \]
   - It suffices to show that \((2^\ell, 2^\ell, \ldots, 2^\ell)\) is attainable from \((1, 2, \ldots, n)\).

7. **Base Case**:
   - For \(n = 3\):
     \[
     (1, 2, 3) \mapsto (2, 2, 4) \mapsto (0, 4, 4) \mapsto (4, 4, 4)
     \]

8. **General Case**:
   - Write \(n = 2^\ell - m\) where \(0 \leq m < 2^{\ell-1}\).
   - If \(m = 0\), we induct on \(n-1\) and add in the \(2^\ell\) already present.
   - Assume \(m \geq 1\). Note that \(m < n\), \(n \equiv m \pmod{2}\), and \(2^{\ell-1}\) is the midpoint of the interval \([m, n]\).
   - Use the operation on the numbers \((2^{\ell-1}-i, 2^{\ell-1}+i)\) for \(1 \leq i \leq n-2^{\ell-1}\):
     \[
     (1, 2, \ldots, m-1, m, m+1, \ldots, n) \mapsto \left(1, 2, \ldots, m-1, 2, 4, \ldots, 2\left(\frac{n-m}{2}\right), 2^{\ell-1}, 2^\ell, 2^\ell, \ldots, 2^\ell\right)
     \]

9. **Casework**:
   - **Case 1**: \(m \geq 4\) and \(n-m \geq 6\):
     - Use inductive hypotheses and scaling on \((1, 2, \ldots, m-1)\) and \((2, 4, 6, \ldots, n-m)\) tuples.
   - **Case 2**: \(m \geq 4\) and \(n-m \leq 4\):
     - Convert \((1, 2, \ldots, m-1) \mapsto (2^{\ell-1}, 2^{\ell-1}, \ldots, 2^{\ell-1})\).
   - **Case 3**: \(m \leq 3\) and \(n-m \geq 6\):
     - Convert \((2, 4, \ldots, n-m) \mapsto (2^\ell, 2^\ell, \ldots, 2^\ell)\).
   - **Case 4**: \(m \leq 3\) and \(n-m \leq 4\):
     - For \(n = 5\):
       \[
       (1, 2, 3, 4, 5) \mapsto (1, 2, 2, 4, 8) \mapsto (1, 0, 4, 4, 8) \mapsto (0, 8, 8, 8, 8) \mapsto (8, 8, 8, 8, 8)
       \]
     - For \(n = 6\):
       \[
       (1, 2, 3, 4, 5, 6) \mapsto (1, 4, 3, 4, 5, 8) \mapsto (1, 4, 2, 4, 8, 8) \mapsto (1, 0, 2, 8, 8, 8) \mapsto (0, 8, 8, 8, 8, 8) \mapsto (8, 8, 8, 8, 8, 8)
       \]

Combining the above cases, we have proven the inductive hypothesis, and consequently, the problem is solved.

The final answer is all powers of two at least \( \boxed{ n } \).