67th Romanian Mathematical Olympiad

The fraction $\frac{x}{y}$ is subunitary, hence $x<y$, that is $x=y-d$, where $d$ is a positive integer. The given relation can be written $\frac{2011-1}{2011}<\frac{y-d}{y}<\frac{2012-1}{2012}$ or $1-\frac{1}{2011}<1-\frac{d}{y}<1-\frac{1}{2012}$, whence $\frac{1}{2011}>\frac{d}{y}>\frac{1}{2012}$ or $\frac{d}{2011d}>\frac{d}{y}>\frac{d}{2012d}$. This leads to $2011d<y<2012d$. (1)

Relation (1) is impossible for $d=1$.

For $d=2$ we get $4022<y<4024$, whence $y=4023$. So $x=4021$ and $x+y=8044$.

For $d\ge 3$, $4021d\ge 12063$. It follows $x+y=2y-d$. From $y>2011d$ follows $2y-d>4021d\ge 12063$.

Consequently, the minimum possible value of the sum is obtained when $d=2$ and $x+y=8044$.