the total number of members is odd. take a member $ x$ notice that there are an even amount of members left. pair them. for any pair $ y,z$ have $ x,y$ give a present to $ z$ and $ x,z$ give a present to $ y$. all presents from $ x$ satisfy the condition now. next we take another member $ q$ and we do the same (*except we have given a present from $ x,q$ already), but now we are left with some $ w$ which is unpaired. put $ q,w$'s present aside and repeat for a different member (not $ w$), again we can pair everyone. now take a different member $ r$, make $ w$ the reminder again. and now pair remainder presents $ qw->r$ and $ rw->q$. so far, every single present from the first 4 members has been given and satisfies the condition. now we see we can repeat this to the end beacuse only the even numbered members leave a remainder present. we also see that this is independent of divisibility by 3, therefore it is all wrong. ? can some1 tell me where i went astray..

Let the children be $ X_i, Y_i, Z_i$ for $ 0 \leq i \leq n-1$; we will be considering the subscripts mod $ n$. The key is that $ 2$ is invertible when $ n$ is odd. Also note that $ (A,B) \to C$ and $ (A,C) \to B$ implies $ (B,C) \to A$, so we basically just need to get each pair of children into a unique triple. First, make the triples $ (X_i, Y_i, Z_i)$ for each $ i$. Then for $ i \neq j$, if $ k$ is equal to $ \frac{i+j}{2}$ in mod $ n$, make the following triples: $ (X_i, X_j, Y_k), (Y_i, Y_j, Z_k), (Z_i, Z_j, X_k)$. This pairs off any children labelled with the same $ X,Y,Z$ letter. Given say $ X_a, Y_b$ (with $ a \neq b$ or the triple is obvious), we know that the third child is $ X_{2b-a}$, and the pair $ Y_b, X_{2b-a}$ also means $ X_a$ was third by the same rule, so the grouping is well defined.

wow somebody actually replied... thank you MellowMelon. im new to olympiad type questions so i am wondering how did you know to look for an answer in specifying the triples instead of say graph theory?

I don't see a good way to get graph theory to work, because what you're dealing with is a function from pairs of kids to kids. Graph theory works better when you have a symmetric relation; I don't see an obvious one here. The triple idea came straight from the observation that $ (A,B) \to C$ implies $ (A,C) \to B$ and $ (B,C) \to A$, so that takes care of the gift given by every pair in the triple $ (A,B,C)$, and that's what suggests grouping them off. The hardest part of the solution was probably just thinking of the $ \frac{i+j}{2} \pmod{n}$ construction. Everything else, like the triple thing, is just toying around with the problem and getting intuition about it. Really, this is how solving most olympiad problems goes. You spend some time developing a basic idea for how things work and find out what makes the problem difficult, and then you try to overcome those barriers.

Asking for the existence of a Steiner triple system of order $ 6n + 3$ (see also http://www.mathlinks.ro/Forum/viewtopic.php?t=240152 ) is soooo original... darij