272 Too 8 people participate in the performance, performing $m$ times, with 4 people on stage each time, requiring any two of the 8 people to be on stage together the same number of times, find the minimum value of $m$.


Suppose each pair of people perform together $k$ times, then all "pairs" appear $\mathrm{C}_{8}^{2} \cdot k=28 k$ times. On the other hand, each performance features $\mathrm{C}_{4}^{2}=6$ "pairs", so the "pairs" appear $6 m$ times in total.
Therefore, $28 k=6 m$, which simplifies to $14 k=3 m$, so $m \geqslant 14$.
An example is as follows, using $1,2, \cdots, 8$ to represent these 8 people:
(1234), (1256), (1278), (1357), (1468), (1367), (1458) , (2368), (2457), (3456), (3478), (5678), (2358), (2467).