Given a natural number $k$, find the smallest natural number $C$ such that
$$\frac C{n+k+1}\binom{2n}{n+k}$$is an integer for every integer $n\ge k$.
1. **Claim and Initial Check:**
   We claim that the smallest natural number \( C \) such that 
   \[
   \frac{C}{n+k+1} \binom{2n}{n+k}
   \]
   is an integer for every integer \( n \ge k \) is \( C = 2k+1 \).

2. **Lower Bound for \( C \):**
   To establish a lower bound for \( C \), consider the case when \( n = k \):
   \[
   \frac{C}{k+k+1} \binom{2k}{k+k} = \frac{C}{2k+1} \binom{2k}{2k} = \frac{C}{2k+1}.
   \]
   Since \( \binom{2k}{2k} = 1 \), the expression simplifies to \( \frac{C}{2k+1} \). For this to be an integer, \( C \) must be divisible by \( 2k+1 \). Therefore, \( C \ge 2k+1 \).

3. **Verification for \( C = 2k+1 \):**
   We need to verify that \( C = 2k+1 \) works for all \( n \ge k \). Consider the binomial coefficient:
   \[
   \binom{2n}{n+k}.
   \]
   We need to show that 
   \[
   \frac{2k+1}{n+k+1} \binom{2n}{n+k}
   \]
   is an integer for all \( n \ge k \).

4. **Expression Simplification:**
   Consider the difference of binomial coefficients:
   \[
   \binom{2n}{n+k} - \binom{2n}{n+k+1}.
   \]
   Using the definition of binomial coefficients, we have:
   \[
   \binom{2n}{n+k} = \frac{(2n)!}{(n+k)!(n-k)!}
   \]
   and
   \[
   \binom{2n}{n+k+1} = \frac{(2n)!}{(n+k+1)!(n-k-1)!}.
   \]
   Therefore,
   \[
   \binom{2n}{n+k} - \binom{2n}{n+k+1} = \frac{(2n)!}{(n+k)!(n-k)!} - \frac{(2n)!}{(n+k+1)!(n-k-1)!}.
   \]

5. **Common Factor Extraction:**
   Factor out the common term \(\frac{(2n)!}{(n+k)!(n-k)!}\):
   \[
   \binom{2n}{n+k} - \binom{2n}{n+k+1} = \frac{(2n)!}{(n+k)!(n-k)!} \left(1 - \frac{(n+k)!(n-k)!}{(n+k+1)!(n-k-1)!}\right).
   \]
   Simplify the fraction inside the parentheses:
   \[
   1 - \frac{(n+k)!(n-k)!}{(n+k+1)!(n-k-1)!} = 1 - \frac{(n-k)}{(n+k+1)} = \frac{2k+1}{n+k+1}.
   \]
   Thus,
   \[
   \binom{2n}{n+k} - \binom{2n}{n+k+1} = \binom{2n}{n+k} \frac{2k+1}{n+k+1}.
   \]

6. **Conclusion:**
   This shows that 
   \[
   \frac{2k+1}{n+k+1} \binom{2n}{n+k}
   \]
   is indeed an integer for all \( n \ge k \). Therefore, the smallest \( C \) that satisfies the condition is \( C = 2k+1 \).

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