Combinatorics problem about finding the possible number of members in a group 
There are $n$ children given. From each two, at least one sends a message to the other. For every child $A$, among the children to each $A$ have sent a message, exactly $10 \%$ sent messages to $A$.
Find all possible 3 digit values for $n$.

This is an interesting combinatorics problem I found in a textbook. I'm not very good at combinatorics and wasn't able to made any significant progress and I decided to ask it here.
Will appreciate any solutions. Thank you in advance!
 A: For a random couple of two children, there is always one message from one child to another, and in $10\%$ of the cases, a message back. The total number of messages thus equals:
$$\frac{11}{10}{n \choose 2} = \frac{11 n (n - 1)}{20}$$
In order for this to result in an integer number of messages, $20$ must divide $n(n - 1)$. However, this condition is not sufficient. Since each child gets exactly $10\%$ of messages back, each child can only send a multiple of $10$ messages. This means that the total number of sent messages must also be a multiple of $10$. Therefore, we find that $200 | 11n(n - 1)$. We can distinguish four cases:


*

*$200 | n$, with four possible values for $n$: $200, 400, 600, 800$

*$200 | n - 1$, with four possible values for $n$: $201, 401, 601, 801$

*$25 | n, 8 | n - 1$, with four possible values for $n$: $225, 425, 625, 825$

*$8 | n, 25 | n - 1$, with five possible values for $n$: $176, 376, 576, 776, 976$
The total number of valid combinations thus equals:
$$4 + 4 + 4 + 5 = 17$$
This can be confirmed using the following Python code:
t = 0
for n in range(100, 1000):
  if n * (n - 1) % 200 == 0:
    t += 1
print(t)

