Junior Balkan Mathematical Olympiad

Solution:
Up to reordering the real numbers $x_{i}$ and $y_{i}$, we may assume that $\frac{x_{1}}{y_{1}} \leq \ldots \leq \frac{x_{n}}{y_{n}}$. Let $A=\frac{x_{1}}{y_{1}}$ and $B=\frac{x_{n}}{y_{n}}$, and $S=\left|x_{1}-y_{1}\right|+\ldots+\left|x_{n}-y_{n}\right|$. Our aim is to prove that $S \leq 2-A-\frac{1}{B}$.

First, note that we cannot have $A>1$, since that would imply $x_{i}>y_{i}$ for all $i \leq n$, hence $x_{1}+\ldots+x_{n}>y_{1}+\ldots+y_{n}$. Similarly, we cannot have $B<1$, since that would imply $x_{i}<y_{i}$ for all $i \leq n$, hence $x_{1}+\ldots+x_{n}<y_{1}+\ldots+y_{n}$.

If $n=1$, then $x_{1}=y_{1}=A=B=1$ and $S=0$, hence $S \leq 2-A-\frac{1}{B}$.

For $n \geq 2$ let $1 \leq k<n$ be some integer such that $\frac{x_{k}}{y_{k}} \leq 1 \leq \frac{x_{k+1}}{y_{k+1}}$. We define the positive real numbers $X_{1}=x_{1}+\ldots+x_{k}$, $X_{2}=x_{k+1}+\ldots+x_{n}$, $Y_{1}=y_{1}+\ldots+y_{k}$, $Y_{2}=y_{k+1}+\ldots+y_{n}$. Note that $Y_{1} \geq X_{1} \geq A Y_{1}$ and $Y_{2} \leq X_{2} \leq B Y_{2}$. Thus, $A \leq \frac{X_{1}}{Y_{1}} \leq 1 \leq \frac{X_{2}}{Y_{2}} \leq B$. In addition, $S=Y_{1}-X_{1}+X_{2}-Y_{2}$.

From $0<X_{2}, Y_{1} \leq 1, 0 \leq Y_{1}-X_{1}$ and $0 \leq X_{2}-Y_{2}$, follows
$$S=Y_{1}-X_{1}+X_{2}-Y_{2}=\frac{Y_{1}-X_{1}}{Y_{1}}+\frac{X_{2}-Y_{2}}{X_{2}}=2-\frac{X_{1}}{Y_{1}}-\frac{Y_{2}}{X_{2}} \leq 2-A-\frac{1}{B}$$