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 $|\frac{m}{n} - (u'+v')| < |\frac{m}{n} - (u+v)|$?
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 $1/2n$.
