Let $U=\{\frac{1}{n}: n\in\mathbb{N}\} \cup \{-\frac{1}{n}: n\in\mathbb{N}\}$ be the set of positive and negative unit fractions.


Are there positive integers $m<n \in \mathbb{N}$, such that for every $u,v\in U$ there are $u',v'\in U$ such that $|\frac{m}{n} - (u'+v')| < |\frac{m}{n} - (u+v)|$?



Approximating rational values in $]0,1[$ by a sum or difference of unit fractions
I suppose the intended inequality is
$$
\left| \frac{m}{n} - |u+v| \, \right| < \left| \frac{m}{n} - |u'+v'| \, \right|
$$
as **Wlodzimierz Holsztynski** suggests. But
there are no such $m,n$. Indeed for any real $r>0$,
rational or not, there exists a unique element of
$$
U + U := \{ u\_1 + u\_2 \mid u\_1,u\_2 \in U \}
$$
closest to $r$.


*Proof*: If $r\in U$ then $r \in U+U$, because
$r = \frac1n = \frac1{2n} + \frac1{2n}$.
If not, choose $n$ such that
$|r| \geq 1/n$. Then if $u\_1,u\_2 \in U$ with
$|r - (u\_1+u\_2)| < \left|r - \frac1n\right|$ then at least one of $u\_1,u\_2$
exceeds $\frac1{2n}$; without loss of generality assume it is $u\_1$.
Then there are finitely many candidates for $u\_1$, and for each of them
there is an optimal choice of $u\_2$ (because $r \neq u\_1$ by hypothesis).
The optimal approximation is then the best one of these as $u\_1$ varies over
the finite list of fractions larger than $1/2n$.


