A company during its vacation managed to organize a table tennis match so that they played one match each day. They also played one game each day on each of the chess sets found in the resort. On the last day of the vacation, they realized that there were no two days when the pairing of chess players was the same, but the next day they could only write down a pairing that had already occurred during the vacation. (In this, they did not consider who played with which color pieces.) How many members were in the company and how long did the vacation last?
Let the number of members of the club, the number of vacation days downloaded, and the number of chess sets be denoted by $t$, $n$, and $s$, respectively. We assume that every member of the club can play both ping-pong and chess. Based on the completion of a ping-pong match,

$$
n=\binom{t}{2}
$$

Considering the chessboards numbered for now, let's write down the number of possible pairings of players to the boards. For the first board, $\binom{t}{2}$ different pairs can be written, for any of them, $\binom{t-2}{2}$ pairs for the second board, and so on, for the $s$-th board, $\binom{t-2(s-1)}{2}$ pairs, so the number of ways to pair all players to the numbered boards is the product of these numbers.

Disregarding the distinguishing numbering of the chessboards, the pairings become identical by a factor of $s!$, so the number of pairing possibilities is smaller by this factor; and, as we know, all of these did indeed occur by the last day, each once, hence

$$
n=\frac{1}{s!} \cdot\binom{t}{2} \cdot\binom{t-2}{2} \cdot\binom{t-4}{2} \cdot \ldots \cdot\binom{t-2 s+2}{2}=\frac{t!}{s!2^{s} \cdot(t-2 s)!}
$$

(It can be assumed that $t-2 s \geqq 0, t \geqq 2 s$, otherwise, it would only have been possible for all $s$ boards to host 1-1 matches daily if some players played more than one game a day; this, however, would need to be specified in the problem statement, without it, the data would be unusable.)

To further transform the equality of the two expressions for $n$, we will aim to get a binomial coefficient on one side and a power of 2 on the other:

$$
\begin{gathered}
\frac{(t-2)!}{s!2^{s-1}(t-2 s)!}=1 \\
\frac{(t-2)!}{(t-2 s)!s!}=\frac{(t-2)!}{(t-2 s)!(2 s-2)!} \cdot \frac{(2 s-2)!}{s!}=2^{s-1} \\
\binom{t-2}{t-2 s} \cdot=(2 s-2) \cdot(2 s-3) \cdot \ldots \cdot(s+2) \cdot(s+1)=2^{s-1}
\end{gathered}
$$

Here, the first factor is also an integer as a binomial coefficient, the others are consecutive integers, and the right side contains only the prime factor 2. Since two consecutive integers include one odd number, this is only possible if the number of additional factors on the left is 1, or there are no such factors, and finally, if there is no factor of 2 on the right, i.e., if $s-1=0, s=1$.

In addition to the first possibility, $2 s-2=s+1$, i.e., $s=3$, and in the second case, $2 s-2=s$, i.e., $s=2$. With these, the following equations are obtained for $t$:

$$
\binom{t-2}{t-6} \cdot 4=4
$$

from which $t-6=0, t=6$; and

$$
\binom{t-2}{t-4}=\binom{t-2}{2}=2, \quad(t-2)(t-3)=4
$$

there is no integer $t$ for this.

The value pair $s=3, t=6$ results in $n=15$, which is a solution to the problem: a 6-member club vacationed for 15 days and conducted 3 chess matches daily (at 3 boards).

Finally, in the case of the last mentioned $s=1$, the roles of the two game types are identical, the chess data does not add anything new to the ping-pong, the problem is indeterminate; otherwise, the problem speaks of sets, so $s>1$.

Note. The difference in pairings does not exclude the possibility that 2 participants play each other more than once, in fact, for $t>4$, it is even desired. For example, $(A B),(C D),(E F)$ and $(A B),(C E),(D F)$ are two different pairings.