Keep those gray cells active!

We're want to compare $\color{#3D99F6} blue$ and $\color{#D61F06} red$ , however, it is easier to compare $\color{#3D99F6} blue$ $+$ $\color{#20A900} green$ and $\color{#D61F06} red$ $+$ $\color{#20A900} green$ .

From similar triangles $\triangle GBJ$ and $\triangle JAC$ , we see $\frac{BJ}{JC}=\frac{GB}{CA}$

And form similar triangles $\triangle EAH$ and $\triangle HCB$ , we see $\frac{CH}{HA}=\frac{BC}{AE}=\frac{GB}{CA}$

Now, we can see that $Area(ABH)=\frac{HA}{CA}*Area(ABC)$ and $Area(JAC)=\frac{JC}{BC}*Area(ABC)$

But $\frac{HA}{CA}=\frac{JC}{BC}$ , so

$Area(ABH)=Area(JAC)$

$\color{#3D99F6} blue$ $+$ $\color{#20A900} green$ $=$ $\color{#D61F06} red$ $+$ $\color{#20A900} green$

$\color{#3D99F6} blue$ $=$ $\color{#D61F06} red$

Call the sides of the squares $a$ and $b$ , respectively.

Use similar triangles $\triangle GFA \sim \triangle JCA$ and $\triangle EDB \sim \triangle HCB$ to conclude $CH = CJ = \frac{ab}{a+b} =: x.$ Now $\text{area}\ \triangle JCA = \tfrac12 b x = \tfrac12 \frac{ab^2}{a+b}; \\ \text{area}\ \triangle BHA = \tfrac12 (b - x) a = \tfrac12 \frac{ab^2}{a+b}.$ Thus $\triangle JCA$ and $\triangle BHA$ have equal areas; subtract $\triangle AHI$ from both to conclude that $\boxed{\text{red}\ =\ \text{blue}}$ .

[The question stated differently when I initially posted it. It said: " ...If area of blue region is 2018, what is the area of red region? " That's why I have 2018 in my calculations.]

First notice $\dfrac{JC}{AC} = \dfrac{BC}{AC+BC}$ and $\dfrac{CH}{BC} = \dfrac{AC}{AC+BC}$ , thus $\dfrac{JC}{CH} = \dfrac{\frac{BC \cdot AC}{AC + BC}}{\frac{AC \cdot BC}{AC+BC}}= 1$ . So, we have $JC = CH$ .

If we denote $P$ as our wanted pink area, we have:

$\begin{aligned} P & = \Delta ABC - \Delta BJA - \Delta ABH + {\color{#3D99F6} \Delta ABI} \\\\ &= \frac{AC \cdot BC}{2} - \frac{AC \cdot BJ}{2} - \frac{BC \cdot AH}{2} + {\color{#3D99F6} 2018} \\\\ &= \frac{AC \cdot BC - AC \cdot (BC - JC) - BC \cdot (AC - CH)}{2} + 2018 \\\\ &= \frac{AC \cdot JC + BC \cdot CH - AC \cdot BC}{2} +2018 ~~~~~~~~~~~~~~~~~~ {\color{#E81990} ~~JC = CH} \\\\ &= \frac{JC \cdot(AC + BC)- AC \cdot BC}{2} + 2018 ~~~~~~~~~~~~~~~~~~~~~~~~~ {\color{#E81990} ~~JC = \dfrac{AC \cdot BC}{AC + BC}} \\\\ &= \frac{\frac{AC \cdot BC}{AC + BC} \cdot(AC + BC)- AC \cdot BC}{2} + 2018 = 2018. \end{aligned}$

Let CA = a and CF=b. Then $\frac{CH}{DE}$ = $\frac{BC}{BD}$ or CH = $\frac{ab}{a+b}$ and area of triangle BCH = $\frac{ab²}{2(a+b)}$ . Likewise, JC= $\frac{ab}{a+b}$ and area of triangle BJA=(a/2)[b- $\frac{ab}{(a+b)}$ ] = $\frac{ab²}{2(a+b)}$ or Area Tr. BCH = Area Tr. BJA or Area Tr. BCH - Area Tr. BJI = Area Tr. BJA - Area Tr. BJI or The Pink Region = The Blue Region no matter what the values a & b are.

Because I say so.

Consider the case when both squares are the same size. By symmetry, $BJ = CJ$ and therefore area $\triangle ACJ =$ area $\triangle ABJ$ . Also by symmetry, area $\triangle BJI =$ area $\triangle AHI$ . Therefore the red and blue regions have the same area.

Scaling $AC$ or $BC$ yields the asymmetric case. But any length scaling leaves the area ratio of the red and blue regions unchanged.

Therefore, the red and blue regions are always equal.

Guess an answer. Like me, you may be right.