Partitioning $[1,v-1]$ into sets of size three such that the sum of each set is $3v/2$ Suppose that $v \equiv 4\;(\bmod 12)$. In general, is it possible to partition the integer interval $[1,v-1]$ into integer partitions of $3v/2$ with three distinct parts (with each part in $[1,v-1]$)? In other words, is it possible in general to partition $[1,v-1]$ into sets of size three such that the sum of each set is $3v/2$?
An exhaustive search for $v = 16$ gives $\{6,8,10\}, \{1,11,12\}, \{2,9,13\}, \{3,7,14\}, \{4,5,15\}$ as a valid partition.
 A: For $v=4$ you only need one set and $\{1,2,3\}$ does the trick.  
For $v=16$ you need five sets summing to $24$.  $\{2,7,15\},\{1,9,14\},\{4,8,12\},\{3,10,11\},\{5,6,13\}$ works.  I found this by splitting the numbers into thirds, small with average $3$, middle with average $8$, and large with average $13$.  Each set has one number from each third.  The numbers within a third were considered to range from $-2$ to $+2$ as the offset from the middle.  I needed five groups that each summed to $0$ and had three of each offset between them.  $-2+1+1$ and $+2-1-1$ suggested themselves along with $\pm2,0$ (twice) and $\pm1,0$.  A little playing.  
For $v=28$ we can again break the numbers into small, medium, and large.  The offsets now range from $-4$ to $+4$ and we need nine sets.  I think the fact that the range is divisible by $3$ makes it easy.  Make three sets with offsets $+4,-1,-3$, rotating, three sets with offsets $-4,+1,+3$, rotating, and three sets with offsets $+2,0,-2$ and we are done.  As the central values are $5,14,23$ the first set is $\{9,13,20\}$ 
I haven't found a general approach.
