In a competition, there were $2n+1$ teams. Every team plays exatly once against every other team. Every match finishes with the victory of one of the teams. We call cyclical a 3-subset of team ${ A,B,C  }$  if $A$ won against $B$, $B$ won against $C$ , $C$ won against $A$.

(a) Find the minimum of cyclical 3-subset (depending on $n$);
(b) Find the maximum of cyclical 3-subset (depending on $n$).
Let's break down the problem into two parts: finding the minimum and maximum number of cyclical 3-subsets.

### Part (a): Minimum Number of Cyclical 3-Subsets

1. **Understanding the Problem**:
   - We have \(2n+1\) teams.
   - Each team plays against every other team exactly once.
   - A cyclical 3-subset \(\{A, B, C\}\) is defined such that \(A\) beats \(B\), \(B\) beats \(C\), and \(C\) beats \(A\).

2. **Graph Representation**:
   - Represent the teams as vertices of a directed graph.
   - Each match result is a directed edge from the winner to the loser.

3. **Minimum Cyclical 3-Subsets**:
   - To minimize the number of cyclical 3-subsets, we need to minimize the number of directed cycles of length 3.
   - One way to achieve this is to have a tournament where the results are as linear as possible, i.e., a directed acyclic graph (DAG).
   - In a DAG, there are no directed cycles, hence no cyclical 3-subsets.

4. **Conclusion**:
   - The minimum number of cyclical 3-subsets is \(0\).

### Part (b): Maximum Number of Cyclical 3-Subsets

1. **Understanding the Problem**:
   - We need to maximize the number of cyclical 3-subsets.

2. **Counting Total 3-Subsets**:
   - The total number of 3-subsets of \(2n+1\) teams is given by:
     \[
     \binom{2n+1}{3} = \frac{(2n+1)(2n)(2n-1)}{6}
     \]

3. **Counting Non-Cyclical 3-Subsets**:
   - A non-cyclical 3-subset is one where one team beats the other two teams.
   - Let \(d_i\) be the number of games that team \(i\) has lost.
   - The number of non-cyclical 3-subsets can be counted as:
     \[
     S = \sum_{i=1}^{2n+1} \binom{d_i}{2}
     \]
   - We know that:
     \[
     \sum_{i=1}^{2n+1} d_i = \binom{2n+1}{2} = n(2n+1)
     \]
   - To minimize \(S\), we need to minimize \(\sum_{i=1}^{2n+1} d_i^2\).

4. **Minimizing \(\sum d_i^2\)**:
   - Using the Cauchy-Schwarz inequality, we have:
     \[
     \left( \sum_{i=1}^{2n+1} 1 \right) \left( \sum_{i=1}^{2n+1} d_i^2 \right) \geq \left( \sum_{i=1}^{2n+1} d_i \right)^2
     \]
   - Substituting the known values:
     \[
     (2n+1) \sum_{i=1}^{2n+1} d_i^2 \geq (n(2n+1))^2
     \]
     \[
     \sum_{i=1}^{2n+1} d_i^2 \geq n^2(2n+1)
     \]

5. **Calculating \(S\)**:
   - The minimum value of \(\sum d_i^2\) is \(n^2(2n+1)\).
   - Therefore:
     \[
     S = \frac{\sum d_i^2 - n(2n+1)}{2} = \frac{n^2(2n+1) - n(2n+1)}{2} = \frac{n(2n+1)(n-1)}{2}
     \]

6. **Calculating Maximum Cyclical 3-Subsets**:
   - The maximum number of cyclical 3-subsets \(T\) is:
     \[
     T = \binom{2n+1}{3} - S
     \]
     \[
     T = \frac{(2n+1)(2n)(2n-1)}{6} - \frac{n(2n+1)(n-1)}{2}
     \]
     \[
     T = \frac{(2n+1)(2n)(2n-1) - 3n(2n+1)(n-1)}{6}
     \]
     \[
     T = \frac{(2n+1)[2n(2n-1) - 3n(n-1)]}{6}
     \]
     \[
     T = \frac{(2n+1)[4n^2 - 2n - 3n^2 + 3n]}{6}
     \]
     \[
     T = \frac{(2n+1)(n^2 + n)}{6}
     \]
     \[
     T = \frac{(2n+1)n(n+1)}{6}
     \]
     \[
     T = \frac{6n^3 + 3n^2 + n}{6}
     \]
     \[
     T = \frac{6n^3 + n^2 - n}{6}
     \]

The final answer is \( \boxed{ \frac{6n^3 + n^2 - n}{6} } \)