Find all positive integers $k$, so that there exists a polynomial $f(x)$ with rational coefficients, such that for all sufficiently large $n$, $$f(n)=\text{lcm}(n+1, n+2, \ldots, n+k).$$
To solve the problem, we need to find all positive integers \( k \) such that there exists a polynomial \( f(x) \) with rational coefficients, which satisfies \( f(n) = \text{lcm}(n+1, n+2, \ldots, n+k) \) for all sufficiently large \( n \).

1. **Understanding the Polynomial Behavior:**
   - Let \( p_\infty \) denote a sufficiently large prime number.
   - Let \( L = \text{lcm}(n+1, n+2, \ldots, n+k) \).
   - We claim that for a polynomial \( P \) with \( \deg(P) \neq 0 \), there exists an \( N \) such that \( P \) is unbounded and strictly increasing or strictly decreasing in \([N, +\infty)\).

2. **Proof by Induction on the Degree of the Polynomial:**
   - **Base Case:** For \( n = 1 \), \( f \) is a linear polynomial, which is either strictly increasing or strictly decreasing and unbounded.
   - **Induction Step:** Assume the claim holds for polynomials of degree \( n \). Consider a polynomial \( P \) of degree \( n+1 \):
     \[
     P(x+1) - P(x) = (a(x+1)^{n+1} + Q(x+1)) - (ax^{n+1} + Q(x))
     \]
     \[
     = a((n+1)x^n + \ldots + 1) + Q(x+1) - Q(x)
     \]
     This is a polynomial of degree \( \leq n \), which by the induction hypothesis is unbounded. Therefore, for sufficiently large \( n \), \( P(x+1) - P(x) \) will be \( > 0 \) or \( < 0 \), implying \( P \) is strictly increasing or decreasing and unbounded.

3. **Behavior of \( f \):**
   - Since \( L > 0 \), \( f \) must be strictly increasing and unbounded.
   - For large \( p_\infty \), we have \( f(p_\infty - 1) < f(p_\infty) \).

4. **Case Analysis for \( k \geq 3 \):**
   - Choose a prime \( q \) such that \( q < k \) and \( q \nmid k \). This is possible because there exists a prime between \( y \) and \( 2y \) (by Bertrand's postulate).
   - By Dirichlet's theorem, we can take \( p_\infty \equiv -k \pmod{q} \).
   - Let \( S = \text{lcm}(p_\infty + 1, p_\infty + 2, \ldots, p_\infty + k - 1) \).
   - Then,
     \[
     f(p_\infty - 1) = \text{lcm}(p_\infty, p_\infty + 1, \ldots, p_\infty + k - 1) = \text{lcm}(p_\infty, S) = \frac{Sp_\infty}{\gcd(p_\infty, S)} = Sp_\infty
     \]
   - Since \( q \mid (p_\infty + k) \) and \( q \mid S \), we have \( \gcd(p_\infty + k, S) \geq q \).
   - Thus,
     \[
     f(p_\infty) = \text{lcm}(p_\infty + 1, p_\infty + 2, \ldots, p_\infty + k) = \text{lcm}(p_\infty + k, S) = \frac{(p_\infty + k)S}{\gcd(p_\infty + k, S)} \leq \frac{(p_\infty + k)S}{q}
     \]
   - Given \( f(p_\infty - 1) < f(p_\infty) \), we get:
     \[
     Sp_\infty < \frac{(p_\infty + k)S}{q} \implies p_\infty(q - 1) < k
     \]
   - This is a contradiction for sufficiently large \( p_\infty \).

5. **Case Analysis for \( k \leq 2 \):**
   - For \( k = 1 \), \( \text{lcm}(n+1) = n+1 \), so \( f(n) = n+1 \).
   - For \( k = 2 \), \( \text{lcm}(n+1, n+2) = \frac{(n+1)(n+2)}{\gcd(n+1, n+2)} = (n+1)(n+2) = n^2 + 3n + 2 \), so \( f(n) = n^2 + 3n + 2 \).

Conclusion:
The only possible values for \( k \) are \( k = 1 \) and \( k = 2 \).

The final answer is \( \boxed{ k = 1 } \) and \( k = 2 \).