partitioning the 3-sets of [n]={1,...,n} into families Let $F_1,...,F_m$ be a partition of the 3-element subsets of $[n]$ into families such that no three subsets in any one family $F_i$ are all contained in one 4-element subset of $[n]$.  What is the minimum value of $m$? 
 A: Tony Huynh's update can be easily generalized to show that
$$n\geq k\left(R(\underbrace{3,3,\dots,3}_{k-1})-1\right)+3\implies m\geq k$$ and so we get a very weak lower bound on $m$ which at least shows that $m _{min}\to \infty$.
For an easy upper bound $m_{min}\le \lfloor\frac{n+1}{2}\rfloor$, which you can see by partitioning the triples $(a,b,c)$ in classes according to $a+b+c\pmod{\lfloor\frac{n+1}{2}\rfloor}$.
A: For a crude lower bound, one can consider the largest possible size for a set in your partition.  One candidate is to take a collection of 2-subsets of $[2n]$,which are triangle-free and then add the point $2n+1$ to each 2-subset.  By Turan's theorem, we get a collection of $n^2$ triples, such that the union of any three is not a 4-set.  This is not best possible, but perhaps it is of the right order. 
Update.
I believe that $m >2$ for all $n \geq 7$. 
Proof.  Towards a contradiction, let $(F_1, F_2)$ be a partition of the 3-sets of $[n]$, such that the union of any three members of $F_i$ is not a 4-set.  Consider the 3-sets of the form $(1,2,k)$.  We may assume that $F_1$ contains at least half of these sets, and since $n \geq 7$, it contains at least 3 sets of this form.  By relabelling, we may assume that $(1,2,3), (1,2,4)$, and $(1,2,5)$ are each in $F_1$.  But this means that $(2,3,4)$, $(2,3,5)$, and $(2,4,5)$ are each not in $F_1$, and hence in $F_2$, a contradiction.  
A: In their investigations into Frankl's union closed sets conjecture, Theresa Vaughn and some of her colleagues considered such configurations of three sets.  I think they were more interested in the size of F_1 than in m.  You might ask her about this problem.
My take on it is that m can be made small, perhaps even m=2, by arranging the 3-sets in cycles.  When I get back to Frankl's problem, I may have more to say.
Gerhard "Ask Me About System Design" Paseman, 2010.10.19
