Autumn tournament

We will first prove the following lemma.

Lemma. Let $X$, $Y$, $Z$ and $T$ be four of the servants. If it is known that the pairs $X,Y$; $Y,Z$; $Z,T$ and $T,X$ were in the room together at some point, then one of the pairs $X,Z$ and $Y,T$ also detected each other.

Proof of Lemma. Let us assume, without community restriction, that $Y$ and $T$ were not in the room together, and $Y$ left the room before $T$ (the other cases are analogous). Then, $X$ and $Z$ were in the room together in the period between $Y$'s departure and $T$'s arrival.

Notice that $A$, $C$, $E$, $F$ satisfy the condition of the lemma, but none of $A,E$ and $C,F$ intersect. The same goes for $A$, $B$, $D$, $G$. The only common element of these pairs is $A$. It remains to be ascertained that it is possible that all the other pairs met, as they claim, by entering exactly once. This is possible with the following sequence of entries and exits: enter $G$, enter $D$, exit $G$, enter $B$, enter $F$, exit $D$, enter $E$, exit $F$, $C$ enters, $B$ exits, $E$ exits, $C$ exits. $■$