Let $n\geq 4$ be a positive integer.Out of $n$ people,each of two individuals play table tennis game(every game has a winner).Find the minimum value of $n$,such that for any possible outcome of the game,there always exist an ordered four people group $(a_{1},a_{2},a_{3},a_{4})$,such that the person $a_{i}$ wins against $a_{j}$ for any $1\leq i<j\leq 4$
1. **Graph Interpretation**: We interpret the problem using a directed graph where each vertex represents a person, and a directed edge from vertex \( u \) to vertex \( v \) indicates that person \( u \) wins against person \( v \).

2. **Total Out-Degree Calculation**: For \( n = 8 \), the total number of directed edges in a complete graph \( K_8 \) is:
   \[
   \binom{8}{2} = 28
   \]
   This means there are 28 directed edges in the graph.

3. **Existence of a Vertex with High Out-Degree**: By the pigeonhole principle, there must be at least one vertex \( a \) with an out-degree of at least:
   \[
   \left\lceil \frac{28}{8} \right\rceil = 4
   \]
   Therefore, there exists a vertex \( a \) that has directed edges to at least 4 other vertices. Let's denote these vertices as \( b, c, d, e \).

4. **Subgraph Analysis**: Consider the subgraph induced by vertices \( b, c, d, e \). This subgraph is a complete directed graph \( K_4 \) with:
   \[
   \binom{4}{2} = 6
   \]
   directed edges.

5. **Existence of a Vertex with High Out-Degree in Subgraph**: Again, by the pigeonhole principle, there must be at least one vertex in this subgraph with an out-degree of at least:
   \[
   \left\lceil \frac{6}{4} \right\rceil = 2
   \]
   Without loss of generality, assume this vertex is \( b \), and it has directed edges to at least two other vertices, say \( c \) and \( d \).

6. **Checking Directed Edges**: 
   - If there is a directed edge \( \overrightarrow{cd} \), then the ordered group \( (a, b, c, d) \) satisfies the condition \( a_i \) wins against \( a_j \) for any \( 1 \leq i < j \leq 4 \).
   - If the directed edge is \( \overrightarrow{dc} \), then the ordered group \( (a, b, d, c) \) satisfies the condition.

   Hence, for \( n = 8 \), there always exists an ordered group of four people satisfying the problem's condition.

7. **Counterexample for \( n = 7 \)**: 
   - Consider the vertices labeled \( 1, 2, 3, 4, 5, 6, 7 \) in \( \mathbb{Z}/7\mathbb{Z} \).
   - Define the directed edge \( \overrightarrow{ab} \) if and only if \( a - b \in \{1, 2, 4\} \).
   - Each vertex has an in-degree and out-degree of 3.
   - Any subset of 4 vertices must form a directed cycle, which contradicts the existence of a directed \( K_4 \) satisfying the problem's condition.

Therefore, \( n = 7 \) does not guarantee the existence of such an ordered group, but \( n = 8 \) does.

The final answer is \( \boxed{8} \)