28. When preparing a district olympiad, each jury member participated in no more than 10 discussions. Discussions can be large or small. In a small discussion, 7 jury members participate, each sending exactly one email to each of the other 6. In a large discussion, 15 jury members participate, each sending exactly one email to each of the other 14. A total of 1994 emails were sent, not counting the emails sent by the jury secretary. In how many discussions did the secretary participate?
28. Answer: The secretary participated in 6 small discussions and 2 large ones.

In a small discussion, 7 jury members participate, each sending exactly one email to each of the 6 others, so in the end, 42 emails are sent in a small discussion. Similarly, participants in a large discussion send 210 emails.

Let a total of $M$ small and $B$ large discussions take place, and the secretary participated in $m \leqslant M$ small and $b \leqslant B$ large discussions, with $m+b \leqslant 10$ by condition. A participant in a small discussion sends 6 emails, and a participant in a large discussion sends 14 emails. Thus, the secretary sent $6 m+14 b$ emails. Adding to this the 1994 emails sent by the other jury members, we get all the emails written as a result of all discussions, i.e., $42 M+210 B$ emails. Therefore,

$$
1994+6 m+14 b=42 M+210 B
$$

The right-hand side of this equation is divisible by 42. Thus,

$$
6 m+14 b \equiv-1994 \equiv 22 \quad(\bmod 42)
$$

Thus, theoretically, the sum $6 m+14 b$ can take values $22, 22+42=64, 22+42 \cdot 2=106$, $22+42 \cdot 3=148$, etc. But since $m+b \leqslant 10$, the sum $6 m+14 b$ cannot be too large:

$$
6 m+14 b \leqslant 14 m+14 b=14(m+b) \leqslant 14 \cdot 10=140 .
$$

Thus, we only need to check whether the equations

$$
6 m+14 b=22, \quad 6 m+14 b=64, \quad 6 m+14 b=106
$$

have solutions in non-negative integers $m$ and $b$, where $m+b \leqslant 10$. This check is performed by a simple enumeration. As a result, we find that the first and third equations do not have such solutions. The second equation has a solution $m=6, b=2$, which seems to suit us.

Now we need to ensure that for the found values of $m$ and $b$, we can choose numbers $M \geqslant m$, $B \geqslant b$ such that the equation $(*)$ is satisfied, i.e.,

$$
42 M+210 B=2058
$$

For example, we can take $B=b=2$, then we find that $M=39$.

Thus, we have found that the situation described in the condition is possible only in the case where the secretary participated in 8 discussions.