First, I am proving that the greatest $\sum \frac 1 {p_i}$ is achieved when $p_i = (2i-1)\cdot 2i$. Suppose that the pair $k, k+1$ is the smallest pair such that $k$ is odd and $k, k+1$ are not in the same subset. Then $k$ is paired with number $a$ and $k+1$ is paired with number $b$ and it holds that $a, b > k+1$ since all numbers less than $k$ are paired consecutively by assumption. Now I will show that we get greater sum by replacing sets $\{k, a\}, \{k+1, b\}$ with $\{k, k+1\}, \{a, b\}$. This is equivalent to $\frac 1{ka}+\frac 1{(k+1)b} < \frac 1{k(k+1)} + \frac 1{ab} \quad \iff \quad k^2+k(1-a-b)+ab-b>0$. The zeroes of the quadratic function $x^2+x(1-a-b)+ab-b$ are $a-1$ and $b$, all of them greater than $k$, therefore the inequality is proven. Now it remains to prove that $\frac 1{1\cdot 2} + \frac 1{3\cdot 4}+\frac 1{5\cdot 6}+\dots <1$. I will prove a stronger inequality $\frac 1{1\cdot 2} + \frac 1{3\cdot 3}+\frac 1{5\cdot 5}+\frac 1{7\cdot 7}+\dots < 1 \quad \iff \frac 1{3^2}+\frac 1{5^2} + \frac 1{7^2} \dots < \frac 1 2$. Denote $x = LHS$ and let $y = \frac 1{2^2}+\frac 1{4^2} + \frac 1{6^2}+ \dots$. It is known that $1+y+x = \frac{\pi^2}6$ and it is easy to notice that $y > x$. Now if $x \ge \frac 1 2$ we get $ \frac{\pi^2}6 = 1+y+x > 2$ which is not true. Therefore $x < \frac 1 2$, QED.

Sorry, for resurrecting an old post. Here's my solution: First show by Rearrangement Inequality that: $ \frac{1}{p_{1}}+\frac{1}{p_{2}}+\ldots+\frac{1}{p_{n}}<\frac {1}{1\cdot 2}+\frac {1}{3\cdot 4}+\frac {1}{5\cdot 6}+\dots+\frac{1}{(2n-1). 2n}$ Now the remaining inequality can be shown in two ways: Method 1: $\frac {1}{1\cdot 2}+\frac {1}{3\cdot 4}+\frac {1}{5\cdot 6}+\dots+\frac{1}{(2n-1). 2n}<\frac {1}{1\cdot 2}+\frac {1}{3\cdot 4}+\frac {1}{5\cdot 6}+\dots\infty=\ln 2\sim 0.693<1$ Method 2: Show by induction:$\frac {1}{1\cdot 2}+\frac {1}{3\cdot 4}+\frac {1}{5\cdot 6}+\dots+\frac{1}{(2n-1). 2n}<\frac{n-1}{n}<1$ Peace. Faustus

Sorry for reviving. Of course the methods shown above actually prove the stronger bound $\log 2<1$. However, if we are only interested in the bound $1$, we can be much more brutal and find a shorter proof without any need for rearrangement: Let $P_i=\{q_i,r_i\}$ where w.l.o.g. $q_i<r_i$ and $q_1<q_2<\dots<q_n$. Then $q_i \ge i$ and $r_i \ge i+1$ and so $p_i \ge i(i+1)$ and the LHS can be bounded by \[\frac{1}{1 \cdot 2}+\frac{1}{2 \cdot 3}+\dots+\frac{1}{n(n+1)}=1-\frac{1}{n+1}<1.\]