Answer
$9 \text{ and } 11$
Work Step by Step
Let $x$ be the odd integer. Then $x+2$ is the other odd integer.
The conditions of the problem translate to
\begin{array}{l}\require{cancel}
\dfrac{1}{x}+\dfrac{1}{x+2}=\dfrac{20}{99}
.\end{array}
Using the properties of equality, the equation above is equivalent to
\begin{array}{l}\require{cancel}
99x(x+2)\left(\dfrac{1}{x}+\dfrac{1}{x+2}\right)=\left(\dfrac{20}{99}\right)99x(x+2)
\\\\
99(x+2)(1)+99x(1)=x(x+2)(20)
\\\\
99x+198+99x=20x^2+40x
\\\\
-20x^2+(99x+99x-40x)+198=0
\\\\
-20x^2+158x+198=0
\\\\
\dfrac{-20x^2+158x+198}{-2}=\dfrac{0}{-2}
\\\\
10x^2-79x-99=0
\\\\
(x-9)(10x+11)=0
.\end{array}
Equating each factor to $0$ (Zero Product Property) and solving for the variable, then the solutions are $x=\left\{ 9,-\dfrac{11}{10}\right\}.$ Since $x$ is an odd integer then $x=9$ is the only acceptable answer. Hence, the odd integers are $
9 \text{ and } 11
.$
