Denote
$$L = \frac{p_1}{p_2 + p_3 + \dots + p_{1005}} + \frac{p_2}{p_3 + p_4 + \dots + p_{1006}} + \dots + \frac{p_{2009}}{p_1 + p_2 + \dots + p_{1004}}$$
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
$$\frac{a_1^2}{b_1} + \frac{a_2^2}{b_2} + \dots + \frac{a_n^2}{b_n} \ge \frac{(a_1 + a_2 + \dots + a_n)^2}{b_1 + b_2 + \dots + b_n}$$
with an equality iff
$$\frac{a_1}{b_1} = \frac{a_2}{b_2} = \dots = \frac{a_n}{b_n}.$$
Then

$$L = \frac{p_1^2}{p_1(p_2 + p_3 + \dots + p_{1005})} + \frac{p_2^2}{p_2(p_3 + p_4 + \dots + p_{1006})} + \dots \\ \qquad + \frac{p_{2009}^2}{p_{2009}(p_1 + p_2 + \dots + p_{1004})}$$
SO
$$L \ge \frac{(p_1 + p_2 + \dots + p_{2009})^2}{\sum_{i=1}^{2009} p_i (p_{i+1} + \dots + p_{i+1004})}.$$
Notice that the denominator of the last fraction is $\sum_{i,j:i<j} p_i p_j$. Let
$$\alpha = \frac{(p_1 + p_2 + \dots + p_{2009})^2}{\sum_{i=1}^{2009} p_i (p_{i+1} + \dots + p_{i+1004})}.$$
Then $(\sum p_i)^2 = \frac{\alpha}{2}((\sum p_i)^2 - (\sum p_i^2))$, so we have
$$\frac{\alpha}{2} \sum p_i^2 = \left(\frac{\alpha}{2} - 1\right) \left(\sum p_i\right)^2 \quad \text{i.e.} \quad \frac{\alpha}{\alpha-2} \left(\sum p_i^2\right) = \left(\sum p_i\right)^2.$$
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
$p_1 = p_2 = \dots = p_{2009}.$