Example 4 A dance and song troupe has $n(n>3)$ actors, who have arranged some performances, with each performance featuring 3 actors performing on stage together. During one performance, they found: it is possible to appropriately arrange several performances so that every 2 actors in the troupe perform on stage together exactly once in this performance. Find the minimum value of $n$. (Original problem)
To solve: Use $n$ points to represent $n$ actors. If two actors have performed together on stage, connect the corresponding points with an edge. Thus, the condition of the problem is equivalent to: the complete graph $K_{n}$ of order $n$ can be partitioned into several complete graphs $K_{3}$ of order 3, such that each edge belongs to exactly one $K_{3}$.

Obviously, $\mathrm{C}_{3}^{2} \mid \mathrm{C}_{n}^{2}$, i.e., $6 \mid n(n-1)$, so $3 \mid n$, or $3 \mid n-1$.
Next, (study other properties of $n$ to narrow down the range), consider the edges containing point $A$ (as a vertex), there are $n-1$ such edges, and each edge belongs to exactly one $K_{3}$, thus there are $n-1$ $K_{3}$s containing point $A$ (as a vertex). But each $K_{3}$ containing point $A$ has 2 edges containing point $A$, so each $K_{3}$ is counted twice, thus $2 \mid n-1$, so $n$ is odd.

From the above, we know that $3 \mid n$ (where $n$ is odd), or $6 \mid n-1$, i.e., $n=6k+3$, or $6k+1\left(k \in \mathbf{N}_{+}\right)$, thus $n \geqslant 7$.

When $n=7$, represent the 7 points with $0,1,2,3,4,5,6$. For $m=0,1,2,3,4,5,6,7$, let $m, m+1, m+3$ form a $K_{3}$ (understand the indices modulo 7, i.e., rotate the $K_{3}$ formed by $0, 1, 3$ 6 times), then the 7 $K_{3}$s form a valid partition.

In summary, the minimum value of $n$ is 7.