Algebra Challenge: Prove $\dfrac{2007}{2}-\ldots=\dfrac{2007}{2008}$

[anemone]

Prove that

$\dfrac{2007}{2}-\dfrac{2006}{3}+\dfrac{2005}{4}-\cdots-\dfrac{2}{2007}+\dfrac{1}{2008}=\dfrac{1}{1005}+\dfrac{3}{1006}+\dfrac{5}{1007}+\cdots+\dfrac{2007}{2008}$

 

[anemone]

Solution proposed by other that I wanted to share it here:

Spoiler

Let the LHS expression be $P$ and the RHS expression be $Q$.

$R=\left(\dfrac{2007}{2}+\dfrac{2006}{3}+\dfrac{2005}{4}+\cdots+\dfrac{1005}{1004}\right)+\left(\dfrac{1004}{1005}+\dfrac{1003}{1006}+\cdots+\dfrac{2}{2007}+\dfrac{1}{2008}\right)$

to both sides, and show that $P+R=Q+R$.

$\begin{align*} P+R&=\left(\dfrac{1}{1005}+\dfrac{1004}{1005}\right)+\left(\dfrac{3}{1006}+\dfrac{1003}{1006}\right)+\cdots+\left(\dfrac{2007}{2008}+\dfrac{1}{2008}\right)+\left(\dfrac{2007}{2}+\dfrac{2006}{3}+\cdots+\dfrac{1005}{1004}\right)\\&=1004+\left(\dfrac{2007}{2}+\dfrac{2006}{3}+\cdots+\dfrac{1005}{1004}\right)\end{align*}$

$\begin{align*} Q+R&=2\left(\dfrac{2007}{2}+\dfrac{2005}{4}+\cdots+\dfrac{1}{2008}\right)\\&=2007+\left(\dfrac{2005}{2}+\dfrac{2003}{3}+\cdots+\dfrac{3}{1003}+\dfrac{1}{1004}\right)\end{align*}$

Subtracting the above from below gives:

$\begin{align*} Q-P&=(2007-1004)+\left(\dfrac{2005-2007}{2}\right)+\left(\dfrac{2003-2006}{3}\right)+\cdots+\left(\dfrac{3-1006}{1003}+\right)+\left(\dfrac{1-1005}{1004}\right)\\&=1003+(-1003)\\&=0\end{align*}$

or simply $P=Q$ and hence we're done.