We have a group of $n$ kids. For each pair of kids, at least one has sent a message to the other one. For each kid $A$, among the kids to whom $A$ has sent a message, exactly $25 \%$ have sent a message to $A$. How many possible two-digit values of $n$ are there?
If the number of pairs of kids with two-way communication is $k$, then by the given condition the total number of messages is $4 k+4 k=8 k$. Thus the number of pairs of kids is $\frac{n(n-1)}{2}=7 k$. This is possible only if $n \equiv 0,1 \bmod 7$.

- In order to obtain $n=7 m+1$, arrange the kids in a circle and let each kid send a message to the first $4 m$ kids to its right and hence receive a message from the first $4 m$ kids to its left. Thus there are exactly $m$ kids to which it has both sent and received messages.
- In order to obtain $n=7 m$, let kid $X$ send no messages (and receive from every other kid). Arrange the remaining $7 m-1$ kids in a circle and let each kid on the circle send a message to the first $4 m-1$ kids to its right and hence receive a message from the first $4 m-1$ kids to its left. Thus there are exactly $m$ kids to which it has both sent and received messages.

There are 26 two-digit numbers with remainder 0 or 1 modulo 7 . (All numbers of the form $7 m$ and $7 m+1$ with $2 \leqslant m \leqslant 14$.)

Alternative Solution by PSC. Suppose kid $x_{i}$ sent $4 d_{i}$ messages. (Guaranteed by the conditions to be a multiple of 4.) Then it received $d_{i}$ messages from the kids that it has sent a message to, and another $n-1-4 d_{i}$ messages from the rest of the kids. So it received a total of $n-1-3 d_{i}$ messages. Since the total number of messages sent is equal to the total number of messages received, we must have:

$$
d_{1}+\cdots+d_{n}=\left(n-1-3 d_{1}\right)+\cdots+\left(n-1-3 d_{n}\right)
$$

This gives $7\left(d_{1}+\cdots+d_{n}\right)=n(n-1)$ from which we get $n \equiv 0,1 \bmod 7$ as in the first solution.

We also present an alternative inductive construction (which turns out to be different from the construction in the first solution).

For the case $n \equiv 0 \bmod 7$, we start with a construction for $7 k$ kids, say $x_{1}, \ldots, x_{7 k}$, and another construction with 7 kids, say $y_{1}, \ldots, y_{7}$. We merge them by demanding that in addition, each kid $x_{i}$ sends and receives gifts according to the following table:

| $i \bmod 7$ | Sends | Receives |
| :---: | :---: | :---: |
| 0 | $y_{1}, y_{2}, y_{3}, y_{4}$ | $y_{4}, y_{5}, y_{6}, y_{7}$ |
| 1 | $y_{2}, y_{3}, y_{4}, y_{5}$ | $y_{5}, y_{6}, y_{7}, y_{1}$ |
| 2 | $y_{3}, y_{4}, y_{5}, y_{6}$ | $y_{6}, y_{7}, y_{1}, y_{2}$ |
| 3 | $y_{4}, y_{5}, y_{6}, y_{7}$ | $y_{7}, y_{1}, y_{2}, y_{3}$ |
| 4 | $y_{5}, y_{6}, y_{7}, y_{1}$ | $y_{1}, y_{2}, y_{3}, y_{4}$ |
| 5 | $y_{6}, y_{7}, y_{1}, y_{2}$ | $y_{2}, y_{3}, y_{4}, y_{5}$ |
| 6 | $y_{7}, y_{1}, y_{2}, y_{3}$ | $y_{3}, y_{4}, y_{5}, y_{6}$ |

So each kid $x_{i}$ sends an additional four messages and receives a message from only one of those four additional kids. Also, each kid $y_{j}$ sends an additional $4 k$ messages and receives from exactly $k$ of those additional kids. So this is a valid construction for $7(k+1)$ kids.

For the case $n \equiv 1 \bmod 7$, we start with a construction for $7 k+1$ kids, say $x_{1}, \ldots, x_{7 k+1}$, and we take another 7 kids, say $y_{1}, \ldots, y_{7}$ for which we do not yet mention how they exchange gifts. The kids $x_{1}, \ldots, x_{7 k+1}$ exchange gifts with the kids $y_{1}, \ldots, y_{7}$ according to the previous table. As before, each kid $x_{i}$ satisfies the conditions. We now put $y_{1}, \ldots, y_{7}$ on a circle and demand that each of $y_{1}, \ldots, y_{3}$ sends gifts to the next four kids on the circle and each of $y_{4}, \ldots, y_{7}$ sends gifts to the next three kids on the circle. It is each to check that the condition is satisfied by each $y_{i}$ as well.