7. A meeting is attended by 24 representatives, and between any two representatives, they either shake hands once or do not shake hands at all. After the meeting, it is found that there were a total of 216 handshakes, and for any two representatives $P$ and $Q$ who have shaken hands, among the remaining 22 representatives, no more than 10 have shaken hands with exactly one of $P$ or $Q$. A "friend circle" refers to a set of three representatives who have all shaken hands with each other. Find the minimum possible number of friend circles among the 24 representatives.
7. Let the 24 representatives be $v_{1}, v_{2}, \cdots, v_{24}$, and for $i=1,2, \cdots, 24$, let $d_{i}$ denote the number of people who have shaken hands with $v_{i}$.
Define the set $E=\left\{\left\{v_{i}, v_{j}\right\} \mid v_{i}\right.$ has shaken hands with $v_{j} \}$.
For any $e=\left\{v_{i}, v_{j}\right\} \in E$, let $t(e)$ be the number of friend circles that include the set $e$.
By the given condition,
$$
\begin{array}{l}
\left(d_{i}-1\right)+\left(d_{j}-1\right) \leqslant 2 t(e)+10 \\
\Rightarrow \sum_{e=\left|v_{i}, v_{j}\right| \in E}\left(d_{i}+d_{j}\right) \leqslant \sum_{e \in E}(2 t(e)+12) .
\end{array}
$$

In $\sum_{e=\mid v_{i}, v_{j} \in E}\left(d_{i}+d_{j}\right)$, the number of handshakes $d_{i}$ of representative $v_{i}$ is counted $d_{i}$ times, so the left side of equation (1) equals $\sum_{i=1}^{24} d_{i}^{2}$. Let the number of friend circles be $T$, noting that each friend circle is counted 3 times in $\sum_{e \in E} t(e)$, and the total number of edges is 216, then the right side of equation (1) equals $6 T+12 \times 216$.
$$
\begin{array}{l}
\text { Hence } \sum_{i=1}^{24} d_{i}^{2} \leqslant 6 T+12 \times 216 . \\
\text { By } \sum_{i=1}^{24} d_{i}^{2} \geqslant \frac{1}{24}\left(\sum_{i=1}^{24} d_{i}\right)^{2}=\frac{(2 \times 216)^{2}}{24} \\
\Rightarrow T \geqslant \frac{1}{6}\left(\frac{(2 \times 216)^{2}}{24}-12 \times 216\right)=864 .
\end{array}
$$

Divide the 24 representatives into four groups, each with 6 people, and let any two representatives from different groups shake hands once, while any two representatives from the same group do not shake hands.
The total number of handshakes is calculated as
$$
\frac{1}{2} \times 24 \times 18=216 \text {. }
$$

For any two representatives $P$ and $Q$ who have shaken hands, among the remaining 22 representatives, exactly 10 representatives have shaken hands with either $P$ or $Q$, which meets the condition. In this case, the number of friend circles is
$$
C_{4}^{3} \times 6^{3}=864 \text {. }
$$

In summary, the minimum possible number of friend circles among the 24 representatives is 864.