Not simply Telescopic

$\begin{aligned} A & = \frac 1{1\times 2} + \frac 1{3\times 4} + \frac 1{5\times 6} + \cdots + \frac 1{99\times 100} \\ & = {\color{#3D99F6}\frac {2-1}{1\times 2}} + {\color{#D61F06}\frac {4-3}{3\times 4}} + {\color{#3D99F6}\frac {6-5}{5\times 6}} + \cdots + {\color{#D61F06}\frac {100-99}{99\times 100}} \\ & = {\color{#3D99F6}\frac 11 - \frac 12} + {\color{#D61F06}\frac 13 - \frac 14} + {\color{#3D99F6}\frac 15 - \frac 16} + \cdots + {\color{#D61F06}\frac 1{99} - \frac 1{100}} \\ & = {\color{#3D99F6}\frac 11 + \frac 12 + \frac 13 + \cdots + \frac 1{100}} - 2 \left(\frac 12 + \frac 14 + \frac 16 + \cdots + \frac 1{100} \right) \\ & = {\color{#3D99F6}H_{100}} - \frac 22 \left(\frac 11 + \frac 12 + \frac 13 + \cdots + \frac 1{50} \right) & \small \color{#3D99F6} \text{where }H_n \text{ is the }n \text{th harmonic number.} \\ & = H_{100} - H_{50} \end{aligned}$

$\begin{aligned} B & = \frac 1{51\times 100} + \frac 1{52\times 99} + \frac 1{53\times 98} + \cdots + \frac 1{100\times 51} \\ & = 2\left(\frac 1{51\times 100} + \frac 1{52\times 99} + \frac 1{53\times 98} + \cdots + \frac 1{75\times 76}\right) \\ & = \frac 2{151} \left(\frac {51+100}{51\times 100} + \frac {52+99}{52\times 99} + \frac {53+98}{53\times 98} + \cdots + \frac {75+76}{75\times 76}\right) \\ & = \frac 2{151} \left(\frac 1{100} + \frac 1{51} + \frac 1{99} + \frac 1{52} + \frac 1{98} + \frac 1{53} + \cdots + \frac 1{76} + \frac 1{75} \right) \\ & = \frac 2{151} \left(H_{100}-H_{50} \right) \end{aligned}$

$\implies \dfrac AB = \dfrac {151}2$ , $\implies m+n = 151+2 = \boxed{153}$ .