Generalize to $2n\times 2n$ points, and consider the center $O$ of the lattice. Consider a polygon $P_1...P_m$. We have by QMGM $$|[P_1...P_m]|\leq \sum_{i=1}^n |[OP_iP_{i+1}]|= \sum_{i=1}^m \frac{1}{2}OP_i\cdot OP_{i+1}\sin(\angle P_iOP_{i+1})\leq \sum_{i=1}^m \frac{1}{4}(OP_i^2+OP_{i+1}^2)=\frac{1}{2}\sum_{i=1}^m OP_i^2$$Summing all these things for the various polygons, we have that the total sum is $$S\leq \frac{1}{2}\sum_{P\in L} OP^2$$. This can be achieved if in all polygons all angles $\angle P_iOP_{i+1}=\frac{\pi}{2}$ and $OP_i=OP_{i+1}$ for all $i$. This means that it can be achieved when and only when we have the set of squares of center $O$. So it finally suffices to compute $\frac{1}{2}\sum_{P\in L} OP^2$, which is easy and left to the reader.

amuthup wrote: Consider a $100\times 100$ square unit lattice $\textbf{L}$ (hence $\textbf{L}$ has $10000$ points). Suppose $\mathcal{F}$ is a set of polygons such that all vertices of polygons in $\mathcal{F}$ lie in $\textbf{L}$ and every point in $\textbf{L}$ is the vertex of exactly one polygon in $\mathcal{F}.$ Find the maximum possible sum of the areas of the polygons in $\mathcal{F}.$ Use a coordinate system so that $\mathbf{L}=\left\{-\frac{99}{2},-\frac{97}{2},-\frac{95}{2},\ldots,\frac{97}{2},\frac{99}{2}\right\}^2$. The area of the polygon $(x_1,y_1),(x_2,y_2).\ldots,(x_n,y_n)$ is $\frac{1}{2}\left|\sum_{cyc}x_iy_{i+1}-x_{i+1}y_i\right|\leq\frac{1}{2}\sum_{cyc}|x_i||y_{i+1}|+|x_{i+1}||y_i|$. Let $A$ be the sum of the areas of the polygons in $\mathcal{F}$ and let $\sigma$ be a permutation of $\mathbf{L}$ that maps each point $(a,b)$ to an adjacent point (in the polygon so that $(a,b)$ is a vertex). By rearrangement \begin{align*} A\leq\frac{1}{2}\sum_{(a,b)\in\mathbf{L}}|a||\mathrm{pr}_Y(\sigma((a,b)))|+|\mathrm{pr}_X(\sigma((a,b)))||b|\leq200\sum_{i=1}^{50}\left(\frac{2i-1}{2}\right)^2 \end{align*}If $\mathcal{F}$ is the set of squares centered at $(0,0)$ which vertices in $\mathbf{L}$ equality holds in every step. Thus $200\sum_{i=1}^{50}\left(\frac{2i-1}{2}\right)^2$ is maximal.

ankan is the god of global! Center the lattice around the origin. The area of each polygon is at most the sum of the areas of the triangles formed by each side of the polygon with the origin, with equality achieving when the origin is inside the polygon. For a polygon, let $\{d_i\}$ denote the distances of its vertices to the origin (in order). It follows that \begin{align*} \sum_{\mathcal P \in \mathcal F}[\mathcal P] &\le \sum_{\mathcal P \in \mathcal F}\sum_{i} \frac{1}{2} d_id_{i+1} \sin{\text{some angle}} \\ &\le \sum_{\mathcal P \in \mathcal F}\sum_{i} \frac{1}{2} d_id_{i+1} \\ &\le \sum_{i=1}^{10000}\frac{1}{2}a_{2i-1}a_{2i}, \end{align*}where $\{a_1, \ldots a_{20000}\}$ is a permutation of all the distances from the $10000$ points to the origin each appearing twice. Rearrangement inequality and extremal principle gives that this quantity is maximized when all $a_{2i-1}$ and $a_{2i}$ are the same. This is achieved when all the polygons are squares centered around the origin. I am feeling too lazy to compute the exact value, so it will be left as an exercise to the reader.

Let the middle of the lattice be $O$. Note that for each polygon $P_1,P_2,\dots, P_n$ in order when considering the ray $OP_i$, we have \begin{align*} A &=[P_1OP_2]+[P_2OP_3]+\dots + [P_nOP_1] \\ &= \sum_{i=1}^{n}{\frac{1}{2}(P_iO)(P_{i+1}O)\sin(\angle P_1)} \\ &\le \sum_{i=1}^{n}{\frac{1}{2}P_iOP_{i+1}O} \\ &\le \sum_{i=1}^{n}{\frac{1}{2}OP_i^2}\end{align*}Extending this across all polygons gives that the sum of the areas is at most half the sum of $OP^2$ for all $P$ in the lattice, which can be achieved with all concentric squares.

The answer is $8,332,500$. This is achieved by having a "spiral of squares", for the outermost layer have the biggest square, and then a square formed by moving each vertex 1 point counterclockwise, and continue doing that until the outer layer is finished. Then, reduce the problem to $n-2$ by $n-2$, and continue from there. We can show by induction that this construction for $2k$ by $2k$ gives a value of $$\frac{k^2(4k^2-1)}{3}$$since we are adding $$(2k+1)^2+(2k)^2\cdots +1^2+1^2+2^2\cdots +(2k)^2$$when going from $k$ to $k+1$. Now, assign a "value" to each vertex, equal to 1/4 of the area of the square that contains it in the above construction described. Let $O$ be the center of the grid (which is not any of the points). Note that since all squares are centered at $O$, the value of a point is also equal to half the square of the distance from the point to $O$. We claim that the area of a polygon is at most the sum of its values of its vertices, which would solve the problem since the sum of all values is $8,332,500$. In fact, we will prove the more general fact that if $O$ is a point in the plane and $A_1\cdots A_n$ is a polygon, then $$[A_1\cdots A_n]\leq \frac{1}{2}(OA_1^2\cdots +OA_n^2).$$Let $$\theta_i=\angle A_iOA_{i+1},$$with indices taken mod $n$. Let $d_i=OA_i.$ Then, we have $$[A_1\cdots A_n]=\sum \frac{1}{2}d_id_{i+1}\sin\theta_i,$$so it suffices to show that $$\sum d_id_{i+1}\sin\theta_i\leq \sum d_i^2.$$This is actually quite a weak inequality, in fact we will show that $$\sum d_id_{i+1}\leq \sum d_i^2,$$which just follows from cyclically summing the inequality $$d_i^2+d_{i+1}^2\geq 2d_id_{i+1}.$$ Remark: another good example of the equality idea, the value assignment is motivated by the equality case with the spiral of squares.

Let $O$ be the average position of all $10000$ points. Then note that by law of cosines and that $n^2$ is a convex function note that the area of each polygon is $\leq$ the sum of the squares from the distance from $O$ to each vertice. Construction for maximum is as follows. The first square is the one made of the vertices of the $99x99$ square. Then the points move right, down, let, up, one unit starting from the top left vertice respectively. Continue this and move the same point until all points on the perimiter of the $99x99$ square has been verticed. Then get rid of this outer square, leaving us with a $97x97$ square, and continue until $1x1.$ By above our final answer must be the sum of the squares of the distance from $O$ to each vertice. Therefore after an fugly bad calculation our answer is $$\boxed{\frac{n^2(2n+1)(2n-1)}{3}}.$$

Note that we can assume $G$ is a tree because we can sum over all trees of $G.$ Consider the highest point in the family tree that is dynastic. Note that only points in the subtree induced by that point can be dynastic otherwise then there is a common ancestor of that point that is also dynastic. Therefore we need only look at a tree such that the root is dynastic. Consider the two children of the root. This induces two new graphs, $K,L.$ We show that the bound is actually $$\frac{n-k-1}{k+2}$$by inductin on $n.$ For base cases $n \leq 2k+3,$ it obviously works as it is $\leq 1,$ so we have that $$\frac{|K|-k-1}{k+2}+\frac{|L|-k-1}{k+2}+1=\frac{|K|+|L|-1-(k+1)}{k+2}<\frac{|G|-k-1}{k+2},$$and as this is tighter than desired we're done$.\blacksquare$

The answer is $\frac 13n^2 (2n-1)(2n+1)$ for a $2n \times 2n$ grid. Construction: We construct inductively. First, draw the square with side length $2n-1$ that encompasses the whole grid. Then, for each $i$, consider the square formed by the two opposite vertices $(i, 0)$ and $(2n-1-i, 2n-1)$ for each $1 \leq i \leq 2n-2$. These $2n-2$ squares have side length $i^2 + (2n-1-i)^2$ each, so the sum of the areas of these squares is equal to $\frac 13(2n-1)(8n^2-8n+3)$. Summing this inductively yields the result. Bound: Let $O$ be the geometric center of the grid. For any polygon $A_1A_2 \cdots A_n$, the area \[[A_1A_2 \cdots A_n] \leq \sum_{i=1}^n OA_i \cdot OA_{i+1} \leq \sum_{i=1}^n \frac 12\left(OA_i^2 + OA_{i+1}^2\right) = \sum_{i=1}^n OA_i^2\]by the sine area formula. Summing this across all points in the grid yields $\sum_{F \in \mathcal F} [F] \leq \sum_{P \in \mathbf L} OP^2$ which is sharp when all the polygons $P$ are squares, as needed. Remark: This is very much a problem where the equality case motivates the solution; what kind of area bound is sharp when we have concentric squares?