Musicians at the Art Festival

Suppose any two musicians co-perform $x$ times. Let the musicians be $A_{1},A_{2},\ldots,A_{8}$ .

Note that the total number of co-performances in each concert is $\binom{4}{2}=6$ . Since there are $k$ concerts, the total number of co-performances is $6k$ .

Next, let $(A_{i},A_{j})$ denote a co-performance of $A_{i}$ and $A_{j}$ , $1\leq i<j\leq 8$ , $i,j \in \mathbb{Z}$ . Note that the total number of different types of co-performances $(A_{i},A_{j})=\binom{8}{2}=28$ because we are choosing $2$ musicians from a total of $8$ musicians. Since any two musicians co-perform $x$ times, the total number of co-performances is $28x$ .

By counting the total number of co-performances in two different ways, we find that $6k=28x \implies 3k=14x$ . Since $3$ and $14$ are coprime and $k$ and $x$ are positive integers, the minimum value of $k$ is $\boxed{14}$ when $x=3$ . Therefore, there are $14$ concerts and any two musicians co-perform $3$ times.

Checking, the following construction works, where $(A_{w},A_{x},A_{y},A_{z})$ means that $A_{w},A_{x},A_{y},A_{z}$ , $1\leq w<x<y<z\leq 8$ , $w,x,y,z \in \mathbb{Z}$ , perform in the same concert: $(A_{1},A_{2},A_{3},A_{4})$ , $(A_{1},A_{2},A_{5},A_{6})$ , $(A_{1},A_{2},A_{7},A_{8})$ , $(A_{1},A_{3},A_{5},A_{7})$ , $(A_{1},A_{3},A_{6},A_{8})$ , $(A_{1},A_{4},A_{5},A_{8})$ , $(A_{1},A_{4},A_{6},A_{7})$ , $(A_{5},A_{6},A_{7},A_{8})$ , $(A_{3},A_{4},A_{7},A_{8})$ , $(A_{3},A_{4},A_{5},A_{6})$ , $(A_{2},A_{4},A_{6},A_{8})$ , $(A_{2},A_{4},A_{5},A_{7})$ , $(A_{2},A_{3},A_{6},A_{7})$ , $(A_{2},A_{3},A_{5},A_{8})$ .