Let $F_1,...,F_m$ be a partition of the 3element subsets of $[n]$ into families such that no three subsets in any one family $F_i$ are all contained in one 4element subset of $[n]$. What is the minimum value of $m$?
Tony Huynh's update can be easily generalized to show that $$n\geq k\left(R(\underbrace{3,3,\dots,3}_{k1})1\right)+3\implies m\geq k$$ 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 $a+b+c\pmod{\lfloor\frac{n+1}{2}\rfloor}$. 


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 2subsets of $[2n]$,which are trianglefree and then add the point $2n+1$ to each 2subset. By Turan's theorem, we get a collection of $n^2$ triples, such that the union of any three is not a 4set. 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 3sets of $[n]$, such that the union of any three members of $F_i$ is not a 4set. Consider the 3sets 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. 


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 3sets 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 


$J(3,n)$
with the least amount of colors such that no triangle is monochromatic. – Moshe Schwartz Oct 19 '10 at 11:24