Let $S$ be the set of all positive rational numbers $r$ such that when the two numbers $r$ and $55r$ are written as fractions in lowest terms, the sum of the numerator and denominator of one fraction is the same as the sum of the numerator and denominator of the other fraction. The sum of all the elements of $S$ can be expressed in the form $\frac{p}{q},$ where $p$ and $q$ are relatively prime positive integers. Find $p+q.$
Denote $r = \frac{a}{b}$, where $\left( a, b \right) = 1$.
We have $55 r = \frac{55a}{b}$.
Suppose $\left( 55, b \right) = 1$, then the sum of the numerator and the denominator of $55r$ is $55a + b$.
This cannot be equal to the sum of the numerator and the denominator of $r$, $a + b$.
Therefore, $\left( 55, b \right) \neq 1$.
Case 1: $b$ can be written as $5c$ with $\left( c, 11 \right) = 1$.
Thus, $55r = \frac{11a}{c}$.
Because the sum of the numerator and the denominator of $r$ and $55r$ are the same,
\[ a + 5c = 11a + c . \]
Hence, $2c = 5 a$.
Because $\left( a, b \right) = 1$, $\left( a, c \right) = 1$.
Thus, $a = 2$ and $c = 5$.
Therefore, $r = \frac{a}{5c} = \frac{2}{25}$.
Case 2: $b$ can be written as $11d$ with $\left( d, 5 \right) = 1$.
Thus, $55r = \frac{5a}{c}$.
Because the sum of the numerator and the denominator of $r$ and $55r$ are the same,
\[ a + 11c = 5a + c . \]
Hence, $2a = 5 c$.
Because $\left( a, b \right) = 1$, $\left( a, c \right) = 1$.
Thus, $a = 5$ and $c = 2$.
Therefore, $r = \frac{a}{11c} = \frac{5}{22}$.
Case 3: $b$ can be written as $55 e$.
Thus, $55r = \frac{a}{c}$.
Because the sum of the numerator and the denominator of $r$ and $55r$ are the same,
\[ a + 55c = a + c . \]
Hence, $c = 0$. This is infeasible.
Thus, there is no solution in this case.
Putting all cases together, $S = \left\{ \frac{2}{25}, \frac{5}{22} \right\}$.
Therefore, the sum of all numbers in $S$ is
\[ \frac{2}{25} + \frac{5}{22} = \frac{169}{550} . \]
Therefore, the answer is $169 + 550 = \boxed{\textbf{(719) }}$.
~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com)