I am looking for a proof for the following result:
Given a convex quadrilateral $ABCD$, divide each of its sides into $n$ segments of equal length (where n is an integer number). Then, connect the corresponding points on opposite sides to form a grid, as shown in the figure then the sum of the areas of the quadrilaterals along a diagonal are equal, and this sum is equal to the $\frac{Area(ABCD)}{n}$. Additionally, the area of the central quadrilateral is equal to the $\frac{Area(ABCD)}{n^2}$.
Up to an affine transformation, without loss of generality the vertices of the quadrilateral are $(0,0),(1,0),(a,b),(0,1)$ for some positive real $a,b$ such that $a+b\ge1$ (so that the quadrilateral be convex). This is because any nonsingular affine transformation changes all areas proportionally.
Now straightforward calculations show that for each $i=0,\dots,n-1$ the area of the $i$th green quadrilateral is $$\frac{2(a+b-2)i+a+b+2 n-2}{2n^3},$$ which is $\ge\frac1{2n^3}>0$, in view of the condition $a+b\ge1$. In particular, for $n=1$ (and $i=0$) this yields the area $\frac{a+b}2$ of the entire big quadrilateral.
Now the desired results easily follow. (The central quadrilateral exists only if $n$ is odd.)
Details: The $j$th quasi-horizontal orange line $m_j$ (through the point $(0,\frac jn)$) has the equation $$y=\frac jn+\frac{\frac jn\,(b-1)}{1+\frac jn\,(a-1)}\,x.$$ Similarly, the $i$th quasi-vertical orange line $l_i$ (through the point $(\frac in,0)$) has the equation $$x=\frac in+\frac{\frac in\,(a-1)}{1+\frac in\,(b-1)}\,y.$$ The point of intersection of the lines $l_i$ and $m_j$ is $$(x_{ij},y_{ij})=\frac1{n^2}\big(i\; ((a-1) j+n),j\; ((b-1) i+n)\big).$$ We also use the expression $$\frac12\det\begin{pmatrix} 1&x_1&y_1\\ 1&x_2&y_2\\ 1&x_3&y_3 \end{pmatrix}\tag{10}\label{10}$$ for the signed area of the triangle with vertices $(x_1,y_1)$, $(x_2,y_2)$, $(x_3,y_3)$ in $\Bbb R^2$.
Now we can clearly see the "reason" for these results: (i) each node $(x_{ij},y_{ij})$ is affine in $(a,b)$ (actually, $x_{ij}$ is an affine function of $a$ and $y_{ij}$ is an affine function of $b$) and (ii) the signed area of a triangle is affine wrt each of the vertices of the triangle. Also, $x_{ij}$ and $y_{ij}$ are affine in $i$ and in $j$.
The calculations can certainly be done manually. However, this is better done using a computer algebra package. Here are details of the calculations with Mathematica:
Most of the ideas of the proof below come from Viktor Prasolov's "Problems in plane and solid geometry". An English translation by Dimitry Leites can be found here. In particular, Problem 4.22 is the statement about the area of the central quadrilateral.
Lemma 1. If $K \in AB$, $L \in BC$, $M \in CD$, $N \in AD$ are points on the sides of a quadrilateral $ABCD$ such that $$\frac{AK}{BK} = \frac{DM}{CM} = \lambda, \quad \frac{AN}{DN} = \frac{BL}{CL} = \mu,$$ then one has $$\frac{NP}{LP} = \lambda, \quad \frac{KP}{MP} = \mu.$$ Proof: Place at the points $A,B,C,D$ masses $1, \lambda, \lambda\mu, \mu$, respectively. Then $K$ is the barycenter of $A$ and $B$ and has mass $1+\lambda$, $M$ is the barycenter of $C$ and $D$ and has mass $\mu(1+\lambda)$. Thus the barycenter of $A, B, C, D$ lies on $KM$ and divides it in the ratio $\lambda : 1$. Similarly, the same barycenter lies on $LN$ and divides it in the ratio $\mu : 1$. $\square$
(Formally, the barycenter is an affine combinations of points.)
Corollary. In the picture in the original post each of the segments joining corresponding points on the opposite sides is divided into five equal parts.
Lemma 2. If $K \in AB$, $L \in BC$, $M \in CD$, $N \in AD$ are the midpoints, then one has $$S(AKPN) + S(CLPM) = S(BKPL) + S(DMPN).$$ Proof: One has $S(AKP) = S(BKP)$ and similarly for three more pairs of triangles. $\square$
Theorem 1. The total area of any five out of 25 quadrilaterals such that no two lie in the same row or in the same column (rook placement) is $1/5$ of the area of the big quadrilateral.
Proof: By Lemmas 1 and 2 in any $2 \times 2$-subcomplex of quadrilaterals the total area of the top left and the bottom right equals the total area of the top right and the bottom left. It follows that there are numbers $a_1, \ldots, a_5$ and $b_1, \ldots, b_5$ such that the area of the quadrilateral on the intersection of the $i$-th row and the $j$-th column equals $a_i + b_j$. Thus for any rook placement the total area is $a_1 + \cdots + a_5 + b_1 + \cdots + b_5$, and the area of the big quadrilateral is five times bigger. $\square$
Lemma 3. If $K, L \in AB$ and $M, N \in CD$ are such that
$$AK = BL = \frac25 AB, \quad CM = DN = \frac25 CD,$$ then one has $S(KLMN) = \frac15 S(ABCD)$.
Proof: One has $$S(ABCD) = |\det(C-A,D-B)|, \quad S(KLMN) = |\det(M-K,N-L)|.$$
At the same time, as $K = \frac35 A + \frac25 B$ etc., one obtains $$M-K = \frac25(C-A) + \frac35(D-B), \quad N-L = \frac35(C-A) + \frac25 (D-B).$$
It follows that $|\det(M-K,N-L)| = \frac15 |\det(C-A,D-B)|$. $\square$
Theorem 2. The area of the central quadrilateral is $1/25$ of the area of the big quadrilateral.
Proof: By Lemma 3, the area of the middle strip is $1/5$ of the area of the big quadrilateral. By Lemma 1, the central quadrilateral sits inside the middle strip in the way described in Lemma 3. Thus its area is $1/5$ of the area of the strip, and hence $1/25$ of the area of the big quadrilateral. $\square$
