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 $x + y = 8044$.