Minimal diameter of set of fractions Let $p_n$ be a pairwise partition of $\{1,2,...,2n\}, n\in \bf N$ where $(a,b)\in p \implies a<b$, and $P_n$ the set of all such pairwise partition. $d(n) := \min_{p_n\in P_n}\big[\max\big(\big|\frac{a}{b}-\frac{c}{d}\big|, (a,b),(c,d)\in p_n\big)\big]$. 
Conjecture: The fraction set from the pairwise partition that achieves $d(n)$ for $n=\frac{3^{i+1}-1}{2}, \,i\in \bf N$ is $\big\{\frac{1}{2},\frac{3}{6},\frac{4}{7},\frac{5}{8}...,\frac{3^i}{2\times3^i},\frac{3^i+1}{2\times 3^i+1},\frac{3^i+2}{2\times3^i+2},...,\frac{2\times3^i-1}{3\times 3^i-1}\big\}$. The other $n$ are constructed similarly.
 A: I believe $(1,2)$ will always be one of the pairs unless $n=2$ when the optimum is $(1,3),(2,4)$ with $d(2)=\frac 16$.  There will always be a fraction at least $\frac 12$, because one of the numerators is at least $n$.  If $1$ is paired with something else, the diameter is at least $\frac 16$  When $2$ or $3$ is used as a denominator, one numerator is at least $n+1$ and that fraction will be strictly greater than $\frac 12$.  It seems easy to keep all the fractions large enough that you would rather have the $\frac 12$ than $\frac 13$
For $n=6$, the best I can do is $(1,2),(3,6),(4,9),(5,10),(7,11),(8,12)$, though you can swap $5,6$ with diameter $\frac 8{12}-\frac 49=\frac 29$  
I think the normal structure of the optimum will be about this:  $(1,2$), some number of fractions that are equal to $\frac 12$, and the rest with all the small numbers in the first position in order and the large ones in the second position in order.  For $n=12$ I get $(1,2),(3,6),(4,8),(5,10),(7,14),(9,18),(10,19),(12,20),(13,21),(15,22),(16,23),(17,24)$ with $d(12)=\frac 5{24}$  This is by hand search-there may be something better out there.  
Following this strategy, assume there are $k$ pairs that equal exactly $\frac 12$.  The numerators will be all the odd numbers, four times the odds, sixteen times the odds, etc.  As $n$ gets large, this is $\frac 23$ of the numbers, so the largest numerator will be about $\frac 32k$.  We then have $n-k$ more pairs, and have used about $\frac k2$ numbers greater than $\frac 32k$, so the last numerator will be about $n+k$.  The first denominator after the halves will be $n+k + 1$  To make sure this fraction is at least $\frac 12$, we need $2(\frac 32k+2)=n+k +1$ or $k=\frac 12(n-3)$.  In the limit of large $n$, we have the final fraction is then $\frac {\frac 32n}{2n}=\frac 34$ and the diameter is $\frac 1{4}$  
Definitely not proven, but I hope some useful thoughts.  An interesting problem.  There could be some improvement by letting the first (few) fractions after the halves drop below.  We saw that in the $n=4$ solution.
