Macedonian Junior Mathematical Olympiad

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

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Alternative solution.

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