An algebra problem about "2019"

$\begin{aligned} S &= 1-2+3-4+5-6+\cdots+2015-2016+2017-2018+2019 \\ &= (1-2)+(3-4)+(5-6)+\cdots+(2015-2016)+(2017-2018)+2019 \\ &= \underbrace{(-1)+(-1)+(-1)}_{1009 \text{ terms}}+2019 \\ &= 1009 \cdot (-1) + 2019 \\ &= -1009+2019 \\ &= \boxed{1010} \end{aligned}$

Another way:

$\begin{aligned} S &= 1-2+3-4+5-6+\cdots+2015-2016+2017-2018+2019 \\ &= {\color{#D61F06}(1+3+5+7+\cdots+2015+2017+2019)} - {\color{#3D99F6}(2+4+6+\cdots+2016+2018)} \\ &= {\color{#D61F06} \left( \frac{2019+1}{2} \right) ^2} - {\color{#3D99F6}2 \cdot (1+2+3+\cdots+1008+1009)} \\ &= {\color{#D61F06}1010^2} - {\color{#3D99F6}2 \cdot \frac {1009 \cdot 1010}2} \\ &= {\color{#D61F06}1010 \cdot 1010} - {\color{#3D99F6}1009 \cdot 1010} \\ &= (1010-1009) \cdot 1010 \\ &= \boxed{1010} \end{aligned}$

$\begin{aligned} 1-2+3-4+\cdots-2018+2019 &= 1+(-2+3)+(-4+5) +\cdots + (-2018+2019) \\ &= 1+\underbrace{1+\cdots+1}_{\frac{2018}{2} \text{ }1\text{s}} \\ &= 1+1009 \\ &= \boxed{1010} \end{aligned}$