Mai's result is :

\(\left(2+4+6+...+98+100\right)-\left(1+3+5+...+97+99\right)\)

\(=\left(2-1\right)+\left(4-3\right)+\left(6-5\right)+...+\left(98-97\right)+\left(100-99\right)\)

\(=1.\left(\dfrac{100-2}{2}+1\right)\) or \(1.\left(\dfrac{99-1}{2}+1\right)\)

\(=50\)