At the end-of-year concert of a music school, four violinists performed. Whenever one of them was not playing, they took a seat among the audience. In at least how many pieces did the violinists perform, if each of them had the opportunity to watch any of their (violinist) colleagues from the auditorium?
We claim that at least 4 performances are necessary for each violinist to hear every other violinist from the audience. First, we show that 3 performances are not sufficient. Let the violinists be denoted by $A, B, C$, and $D$. It is not possible for each performance to feature only one violinist, as one would be left out. Therefore, at least two violinists must perform together in one of the performances. Since there are a total of 3 performances, two cases are possible:

1. Each violinist performs once and sits in the audience twice;
2. There is a violinist - let's say $A$, who performs twice and sits in the audience once.

In the first case:

performs

watches

1. performance

$A$

$B, C, D$

2. performance

$B, C$

$A, D$

$A, B, C$

However, in this case, the two violinists ($B$ and $C$) who performed together did not hear each other. (This does not depend on the choice of letters).

In the second case, if $A$ performs twice, then $A$ can only sit in the audience once. The other two performances must be arranged so that the others also sit in the audience at least once. It is not possible for only one performer to sit in the audience, because there are only two more performances, and no one can sit in both, as the others would not be able to see. For example:

performs

watches

| 1. performance | 2. performance | 3. performance |
| :---: | :---: | :---: |
| $A$ | $A$ | $B, C, D$ |
| $B, C$ | $D$ | $A$ |

But in this case, $B$ and $C$ did not hear each other either.

However, four performances are certainly sufficient, for example, as follows:

performs

watches

1. performance

$A$

$B, C, D$

2. performance

$B$

$A, C, D$

3. performance

$C$

$A, B, D$

4. performance

$D$

$A, B, C$