Solution:

If the number of pairs of kids with two-way communication is $k$, then by the given condition the total number of messages is $4k + 4k = 8k$. Thus the number of pairs of kids is $\frac{n(n-1)}{2} = 7k$. This is possible only if $n \equiv 0,1 \pmod{7}$.

- In order to obtain $n = 7m + 1$, arrange the kids in a circle and let each kid send a message to the first $4m$ kids to its right and hence receive a message from the first $4m$ kids to its left. Thus there are exactly $m$ kids to which it has both sent and received messages.

- In order to obtain $n = 7m$, let kid $X$ send no messages (and receive from every other kid). Arrange the remaining $7m - 1$ kids in a circle and let each kid on the circle send a message to the first $4m - 1$ kids to its right and hence receive a message from the first $4m - 1$ 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 $7m$ and $7m+1$ with $2 \leqslant m \leqslant 14$.)

Solution:

Suppose kid $x_{i}$ sent $4d_{i}$ 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 $n-1-4d_{i}$ messages from the rest of the kids. So it received a total of $n-1-3d_{i}$ messages. Since the total number of messages sent is equal to the total number of messages received, we must have:
$$d_{1} + \cdots + d_{n} = (n-1-3d_{1}) + \cdots + (n-1-3d_{n})$$
This gives $7(d_{1} + \cdots + d_{n}) = n(n-1)$ from which we get $n \equiv 0,1 \pmod{7}$ 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 $n \equiv 0 \pmod{7}$, we start with a construction for $7k$ kids, say $x_{1}, \ldots, x_{7k}$, 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:

For the case $n \equiv 1 \pmod{7}$, we start with a construction for $7k+1$ kids, say $x_{1}, \ldots, x_{7k+1}$, 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 $x_{1}, \ldots, x_{7k+1}$ 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 easy to check that the condition is satisfied by each $y_{i}$ as well.