Macedonian Junior Mathematical Olympiad

For arbitrary natural numbers $n \neq 1 \neq m$ the inequality $nm \geq n+m$ holds, since $(n-1)(m-1) \geq 1 \Rightarrow nm-n-m+1 \geq 1 \Rightarrow nm-n-m \geq 0 \Rightarrow nm \geq n+m$, with equality only when $n=m=2$. Then for $n \geq 2$, we have
$$\frac{1}{n(2014-n)} < \frac{1}{n+2014-n} = \frac{1}{2014}$$
and hence:
$$\frac{1}{1 \cdot 2013} + \frac{1}{2 \cdot 2012} + \frac{1}{3 \cdot 2011} + \dots + \frac{1}{2012 \cdot 2} + \frac{1}{2013 \cdot 1} < \frac{2}{2013} + \frac{2011}{2014} < \frac{3}{2014} + \frac{2011}{2014} = 1.$$

$$\begin{aligned} & \frac{1}{1 \cdot 2013} + \frac{1}{2 \cdot 2012} + \frac{1}{3 \cdot 2011} + \dots + \frac{1}{2012 \cdot 2} + \frac{1}{2013 \cdot 1} = \\ &= \frac{1}{2014} \left( \frac{1+2013}{1 \cdot 2013} + \frac{2+2012}{2 \cdot 2012} + \frac{3+2011}{3 \cdot 2011} + \dots + \frac{2012+2}{2012 \cdot 2} + \frac{2013+1}{2013 \cdot 1} \right) = \\ &= \frac{1}{2014} \left( \left( \frac{1}{1} + \frac{1}{2013} \right) + \left( \frac{1}{2} + \frac{1}{2012} \right) + \dots + \left( \frac{1}{2012} + \frac{1}{2} \right) + \left( \frac{1}{2013} + \frac{1}{1} \right) \right) = \\ &= \frac{1}{2014} \left( 2 \left( \frac{1}{1} + \frac{1}{2} + \dots + \frac{1}{2012} \right) + \frac{1}{2013} \right) = \frac{1}{2014} \left( 3 + \frac{2}{3} + \frac{2}{4} + \dots + \frac{2}{2012} + \frac{2}{2013} \right) < \\ &< \frac{1}{2014} \left( 3 + \frac{1+1+\dots+1+1}{2014} \right) = 1 \end{aligned}$$