Is Brute Force the only way?

$20=4\cdot 5, 30=5\cdot 6,\cdots ,90=9\cdot 10$ The sum $(\mathfrak T)$ is : $\Large{\begin{aligned}\mathfrak{T}=&\displaystyle\sum_{r=4}^{9}\dfrac{1}{r(r+1)}\\=&\displaystyle\sum_{r=4}^{9}\left(\dfrac{1}{r}-\dfrac{1}{r+1}\right)\end{aligned}}$

$\color{#0C6AC7}{\mathbf{(A~Telescopic~Series)}}$ $\Large =\left(\dfrac{1}{4}-\dfrac{1}{10}\right)=\dfrac{3}{20}$

$\huge\therefore 20+3=\color{#D61F06}{\mathbf{\boxed{\color{#20A900}{23}}}}$