We say that a group of $k$ boys is $n-acceptable$ if removing any boy from the group one can always find, in the other $k-1$ group,  a group of $n$ boys such that everyone knows each other. For each $n$, find the biggest $k$ such that in any group of $k$ boys that is $n-acceptable$ we must always have a group of $n+1$ boys such that everyone knows each other.
To solve this problem, we need to find the largest integer \( k \) such that any \( k \)-boy group that is \( n \)-acceptable must contain a group of \( n+1 \) boys who all know each other. We will use the concept of complete graphs and the properties of \( n \)-acceptable groups to derive the solution.

1. **Define \( n \)-acceptable group**:
   A group of \( k \) boys is \( n \)-acceptable if removing any boy from the group leaves a group of \( k-1 \) boys in which there is always a group of \( n \) boys who all know each other.

2. **Consider the function \( f(A) \)**:
   For a group \( A \) of \( 2n-1 \) boys, let \( f(A) \) be the number of complete \( n \)-graphs (i.e., groups of \( n \) boys who all know each other) within that group.

3. **Minimal counterexample**:
   Assume there is a counterexample for some \( n \). Choose the counterexample with the minimal \( f(A) = k \). This means that in this group of \( 2n-1 \) boys, there are exactly \( k \) complete \( n \)-graphs.

4. **Non-empty intersection property**:
   For any \( k-1 \) complete \( n \)-graphs, there must be a non-empty intersection. This is because if there were no intersection, we could remove one boy and still find a complete \( n \)-graph, contradicting the minimality of \( f(A) \).

5. **Contradiction and reduction**:
   Using the non-empty intersection property, we can show that either the configuration is not a counterexample, or there exists a counterexample with \( f(A) = k-1 \). This leads to a contradiction because we assumed \( f(A) \) was minimal.

6. **Conclusion**:
   By the properties of \( n \)-acceptable groups and the minimal counterexample argument, we can conclude that the largest \( k \) such that any \( k \)-boy group that is \( n \)-acceptable must contain a group of \( n+1 \) boys who all know each other is \( k = 2n-1 \).

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