<s>[INST] The Arts and Crafts of Problem Solving Problem 2.2.20 Let $n$ be a positive integer.
We can find integers $a,b$ such that $1 \leq a < b \leq n$, $a+b > n$ and $a,b$ are relatively prime.
Then we find $\frac{1}{ab}$.
Then we sum all of the found fractions.
For example if $n=6$, the ordered pairs are 
$
(1,6),(5,6),(2,5),(3,5),(4,5),(3,4)
$
and the corresponding sum is 
$
\frac{1}{6} + \frac{1}{30} + \frac{1}{10} + \frac{1}{15} + \frac{1}{20} + \frac{1}{12} = \frac{1}{2}
$
Investigate what happens with the other values of $n$, and conjecture something.
What I have found so far is that the sum always stays the same but I am having issues proving that it is true.
$\textbf{Partial Solution}$
If we start experimenting with more values of $n$, such as $n = 2$. Then the ordered pairs are
(1,2)
Then the corresponding sum is 
$
\frac{1}{2} = \frac{1}{2}
$
$n = 5$. Then the ordered pairs are
(1,5),(2,5),(3,5),(4,5),(3,4)
Then the corresponding sum is 
$
\frac{1}{5} + \frac{1}{10} + \frac{1}{15} + \frac{1}{20} + \frac{1}{12} = \frac{1}{2}
$
From this we can assume that the sum stays the same for all $n$
 in the positive integers
 [/INST] (My deleted solution shows more details.)
Hint: Induction.
Consider going from $n $ to $ n+1 $. (Use $n = 6$ to help think it through).   
Step 1. Completely categorize all the pairs that you lose.
Step 2. Completely categorize all the pairs that you gain.
Step 3. Show that the sum from terms that you lose is equal to the sum from terms that you gain.
Step 4: Hence, the sum is a constant. 
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