Given positive integer $n$, find the biggest real number $C$ which satisfy the condition that if the sum of the reciprocals of a set of integers (They can be the same.) that are greater than $1$ is less than $C$, then we can divide the set of numbers into no more than $n$ groups so that the sum of reciprocals of every group is less than $1$.
1. **Identify the problem and the goal:**
   We need to find the largest real number \( C \) such that if the sum of the reciprocals of a set of integers greater than 1 is less than \( C \), then we can divide the set into no more than \( n \) groups, each with a sum of reciprocals less than 1.

2. **Propose a candidate for \( C \):**
   We propose \( C = \frac{n+1}{2} \). To justify this, we need to show that if the sum of the reciprocals is less than \( \frac{n+1}{2} \), then the set can be divided into \( n \) groups with each group having a sum of reciprocals less than 1.

3. **Construct a counterexample for \( C = \frac{n+1}{2} \):**
   Consider \( n+1 \) instances of \( \frac{1}{2} \). The sum of these reciprocals is:
   \[
   \sum_{i=1}^{n+1} \frac{1}{2} = \frac{n+1}{2}
   \]
   If we try to divide these \( n+1 \) instances into \( n \) groups, at least one group will contain at least two instances of \( \frac{1}{2} \), making the sum of reciprocals in that group equal to 1, which violates the condition.

4. **Verify the upper bound:**
   Now, we need to show that if the sum of the reciprocals is less than \( \frac{n+1}{2} \), we can indeed divide the set into \( n \) groups with each group having a sum of reciprocals less than 1.

5. **Proof by contradiction:**
   Suppose we have a set of integers whose sum of reciprocals is less than \( \frac{n+1}{2} \), and we cannot divide them into \( n \) groups with each group having a sum of reciprocals less than 1. This means that at some point, we fail to place \( \frac{1}{h} \) into any group because the sum in any group is at least \( 1 - \frac{1}{h} \).

6. **Calculate the total sum:**
   If each group has a sum of at least \( 1 - \frac{1}{h} \), then the total sum of the reciprocals in all groups is at least:
   \[
   n \left(1 - \frac{1}{h}\right) = n - \frac{n}{h}
   \]
   Since \( h \) is an integer greater than 1, the smallest value for \( \frac{n}{h} \) is \( \frac{n}{2} \) when \( h = 2 \). Thus, the total sum is at least:
   \[
   n - \frac{n}{2} = \frac{n}{2}
   \]
   However, this contradicts our assumption that the sum of the reciprocals is less than \( \frac{n+1}{2} \).

7. **Conclusion:**
   Therefore, if the sum of the reciprocals is less than \( \frac{n+1}{2} \), we can always divide the set into \( n \) groups with each group having a sum of reciprocals less than 1.

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