find all $k$ distinct integers $a_1,a_2,...,a_k$ such that there exists an injective function $f$ from reals to themselves such that for each positive integer $n$ we have
$$\{f^n(x)-x| x \in \mathbb{R} \}=\{a_1+n,a_2+n,...,a_k+n\}$$.
1. **Claim 1: \(a_1, a_2, \ldots, a_k \geq 0\)**

   **Proof:** Suppose for the sake of contradiction that \(a_1\) is the minimum and \(a_1 < 0\). Let \(a = -a_1\). Then, for some \(x\), we have:
   \[
   f^a(x) = x + a + a_1 = x \implies f^a(x) = x
   \]
   This implies that \(f\) has a periodic point with period \(a\). Now, for some \(a_j\), we have:
   \[
   f^{2a}(x) = f^a(f^a(x)) = f^a(x) = x \implies x + 2a + a_j = x \implies -2a = a_j
   \]
   But we had \(a_1 < 0\) as the minimum, leading to \(2a_1 = a_j\), which is a contradiction. Hence, \(a_1, a_2, \ldots, a_k \geq 0\).

2. Let \(M > 0\) be the maximum number in \(\{a_1, \ldots, a_k\}\). There exists some \(x\) such that \(f(x) = x + M + 1\). Then:
   \[
   f^2(x) = x + 2 + a_i \quad \text{for some } i
   \]
   Also:
   \[
   f^2(x) = f(x + M + 1) = x + M + 2 + a_j \quad \text{for some } j
   \]
   This implies:
   \[
   a_j + M = a_i
   \]
   Since \(M\) is the maximum and \(a_1, \ldots, a_k \geq 0\), we have \(a_j = 0\) and \(a_i = M\). Thus:
   \[
   f^2(x) = x + M + 2
   \]

3. Similarly, if for \(i = 1, \ldots, l\), \(f^l(x) = x + M + l\), then:
   \[
   f^{l+1}(x) = f(f^l(x)) = f(x + M + l) = x + M + l + 1 + a_i
   \]
   Also:
   \[
   f^{l+1}(x) = x + l + 1 + a_j
   \]
   So we get:
   \[
   M + l + 1 + a_i = l + 1 + a_j \implies M + a_i = a_j \implies a_i = 0, a_j = M
   \]
   Therefore, for all \(j\), \(f^j(x) = x + j + M\).

4. Let \(f(x + M) = x + M + 1 + a_i\). Then:
   \[
   f^{a_i + 1}(x) = f(x + M)
   \]
   If \(a_i > 0\), then:
   \[
   f^{a_i}(x) = x + M \implies x + a_i + M = x + M \implies a_i = 0
   \]
   This is a contradiction. If \(a_i = 0\), then:
   \[
   f(x + M) = x + M + 1 = f(x) \implies x = x + M \implies M = 0
   \]
   This is also a contradiction.

5. Therefore, we must have \(M = 0\), so the only possible set is \(\{0\}\). We need \(f^i(x) - x = i\). Notice that \(f(x) = x + 1\) works fine.

\(\blacksquare\)

The final answer is \( \boxed{ \{0\} } \)