Largest subset which contains no three equal sums. What is the largest possible size a subset $X$ of $[1,899]$ can have such that there are no three distinct subsets of $X : \{a,b\}\{c,d\}\{e,f\}$ such that $a+b=c+d=e+f$
 A: Not an answer, but here are lower and upper bounds.  For a crude upper bound, an $n$-subset $X$ of $[1,899]$ contains $\binom{n}{2}$ possible sums of $2$-sets.  There are at most $1895$ possible sums.  Thus, if $\binom{n}{2} \geq 3791$, $X$ will contain $3$ equal sums.  This happens for $n \geq 88$.
Upper bound: $87$.
For the lower bound one can take the elements of the sequence
$1,2,3,4,5,7,9,12,16,21,28,37,49, 65, \dots$
This sequence is defined recursively by $a_i=i$ for $i=1,\dots, 5$ and then setting $a_n:=\min(a_{n-1}+a_{n-4}, a_{n-2}+a_{n-3})$.  With this choice, no $a_n$ can be the largest element of a valid $6$-tuple, since it is too large to be so.  This sequence has $23$ elements before it becomes larger than $899$.
Lower bound: $23$.   
Here is a different construction that gives a slightly bigger set.  Consider the sequence
$1,2,4, 8,9,11, 20,21,23, 44, 45, 47, 92, 93, 95, 188, 189, 191, 380, 381, 383, 764, 765, 767$
This sequence is naturally partitioned into blocks of size $3$.  I claim that there does not exist a valid triple $\{a,b\}, \{c,d\}, \{e,f\}$.  Let $B_1$ be the last block which intersects $T:=\{a,b,c,d,e,f\}$. Each of $\{a,b\}, \{c,d\}, \{e,f\}$ must intersect $B_1$ since the first member of $B_1$ is as large as any pair coming earlier in the sequence.  But in this sequence, blocks are the only triples that have the difference set $\{1,2,3\}$. Thus, $T$ also intersects another block $B_2$ completely. But it is easy to see that $T$ cannot be the union of two blocks.  
New lower bound: $24.$ 
