Three, (50 points) A dance troupe has $n(n \geqslant 5)$ actors, and they have arranged some performances, each of which is performed by four actors on stage. In one performance, they found that: it is possible to appropriately arrange several performances so that every two actors in the troupe perform on stage together exactly once during this performance. Find the minimum value of $n$.

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Three, use $n$ points to represent $n$ actors.
If two actors have performed on the same stage once, then connect the corresponding points with an edge. Thus, the condition of this problem is equivalent to:

Being able to partition the complete graph $K_{n}$ of order $n$ into several complete graphs $K_{4}$ of order 4, such that each edge belongs to exactly one $K_{4}$.
First, from the problem statement, we know $\mathrm{C}_{4}^{2} I \mathrm{C}_{n}^{2}$, i.e.,
$12 \ln (n-1)$.
Therefore, $n \neq 5, 6, 7, 8, 10, 11$.
Second, consider the edges containing point $A$ (as an endpoint), there are $n-1$ such edges, and each edge belongs to exactly one $K_{4}$, thus, there are $n-1$ $K_{4}$s containing point $A$ (as an endpoint). However, each $K_{4}$ containing point $A$ has three edges containing point $A$, thus, each $K_{4}$ is counted 3 times.
Thus, $31(n-1)$.
Therefore, $n \neq 9, 12$.
Finally, represent 13 points using $0,1, \cdots, 12$. For $m=0,1, \cdots, 12$, let $m, m+1, m+5, m+11$ form a $K_{4}$ (points are considered modulo 13). Then the 13 $K_{4}$s form a valid partition.
In summary, the minimum value of $n$ is 13.