North Macedonia

Denote $$L=\frac{p_1}{p_2+p_3+\cdots +p_{1005}}+\frac{p_2}{p_3+p_4+\cdots +p_{1006}}+\cdots +\frac{p_{2009}}{p_1+p_2+\cdots +p_{1004}}$$ We will use the Cauchy-Schwarz inequality in the form: For every $a_1,a_2,\dots ,a_n\in \mathbb{R}$, $b_1,b_2,\dots ,b_n\in {\mathbb{R}}^{+}$ it holds $$\frac{a_1^2}{b_1}+\frac{a_2^2}{b_2}+\cdots +\frac{a_n^2}{b_n}\ge \frac{(a_1+a_2+\cdots +a_n{)}^2}{b_1+b_2+\cdots +b_n}$$ with an equality iff $$\frac{a_1}{b_1}=\frac{a_2}{b_2}=\cdots =\frac{a_n}{b_n}.$$ Then

$$\begin{gathered} L=\frac{p_1^2}{p_1(p_2+p_3+\cdots +p_{1005})}+\frac{p_2^2}{p_2(p_3+p_4+\cdots +p_{1006})}+\dots \\ +\frac{p_{2009}^2}{p_{2009}(p_1+p_2+\cdots +p_{1004})} \end{gathered}$$ SO $$L\ge \frac{(p_1+p_2+\cdots +p_{2009}{)}^2}{{\sum }_{i=1}^{2009}{p}_i(p_{i+1}+\cdots +p_{i+1004})}.$$ Notice that the denominator of the last fraction is ${\sum }_{i,j:i<j}{p}_i{p}_j$. Let $$\alpha =\frac{(p_1+p_2+\cdots +p_{2009}{)}^2}{{\sum }_{i=1}^{2009}{p}_i(p_{i+1}+\cdots +p_{i+1004})}.$$ Then $(\sum p_i{)}^2=\frac{\alpha }{2}((\sum p_i{)}^2-(\sum p_i^2))$, so we have $$\frac{\alpha }{2}\sum p_i^2=\left(\frac{\alpha }{2}-1\right){\left(\sum p_i\right)}^2\text{i.e.}\frac{\alpha }{\alpha -2}\left(\sum p_i^2\right)={\left(\sum p_i\right)}^2.$$ So $0<\frac{\alpha }{\alpha -2}\le 2009$ (we get the second inequality from Cauchy-Schwarz or the inequality between the quadratic and arithmetic mean). This inequality is actually equivalent to $\alpha \ge \frac{2009}{1004}$. Thus, $L\ge \frac{2009}{1004}$ and equality holds iff $p_1=p_2=\cdots =p_{2009}.$