, 2022

The maximum possible value of the sum of the areas is $\Sigma(n) = \frac{1}{3}n^2(2n+1)(2n-1)$.

Below, let $[P]$ denote the area of the polygon $P$. We first make some conventions. If all the vertices of a polygon belong to $S$, we call it legal. Let $O = (n - \frac{1}{2}, n - \frac{1}{2})$ be the center point of $S$. If a legal square has center $O$, we call that square centered. Finally, if a set of polygons $\mathcal{F}$ satisfies the conditions of the problem, we call $\mathcal{F}$ admissible, and denote the sum of its areas by $\Sigma(\mathcal{F})$.

Every point in $S$ is exactly the vertex of a unique centered square. So all the centered squares form an admissible set $\mathcal{G}$. Below we prove that
$$\Sigma(\mathcal{F}) \le \Sigma(\mathcal{G}) = \Sigma(n) \qquad (1)$$
holds for every admissible $\mathcal{F}$, which thus answers the problem.

We use the following key lemma.

Lemma 1. Let $P = A_1A_2\dots A_m$ be a polygon, and let $O$ be an arbitrary point in the plane. Then
$$[P] \le \frac{1}{2} \sum_{i=1}^{m} OA_i^2; \qquad (2)$$
and moreover, when $P$ is a square and $O$ is its center, the inequality in (2) becomes an equality.

Proof of Lemma. Let $P_{n+1} = P_1$. For each $i = 1, 2, \dots, m$, we have
$$[OA_iA_{i+1}] \le \frac{OA_i \cdot OA_{i+1}}{2} \le \frac{OA_i^2 + OA_{i+1}^2}{4}.$$
Thus we obtain
$$[P] \le \sum_{i=1}^{m} [OA_iA_{i+1}] \le \frac{1}{4} \sum_{i=1}^{m} (OA_i^2 + OA_{i+1}^2) = \frac{1}{2} \sum_{i=1}^{m} OA_i^2,$$
which is exactly (2). Direct verification shows that when $P$ is a square and $O$ is its center, the inequality in (2) becomes an equality. This proves the lemma. $\square$

$$\Sigma(\mathcal{F}) \le \frac{1}{2} \sum_{A \in \mathcal{S}} OA^2 = \Sigma(\mathcal{G}),$$
so the left-hand inequality of (1) holds.

Finally, we compute $\Sigma(\mathcal{G})$. From (2) we know
$$\begin{align*} \Sigma(\mathcal{G}) &= \frac{1}{2} \sum_{A \in \mathcal{S}} OA^2 = \frac{1}{2} \sum_{i=0}^{2n-1} \sum_{j=0}^{2n-1} \left( \left( n - \frac{1}{2} - i \right)^2 + \left( n - \frac{1}{2} - j \right)^2 \right) \\ &= \frac{1}{8} \cdot 4 \cdot 2n \sum_{i=0}^{n-1} (2n - 2i - 1)^2 \\ &= n \sum_{j=0}^{n-1} (2j + 1)^2 \\ &= \frac{n^2(2n + 1)(2n - 1)}{3} = \Sigma(n). \end{align*}$$
This completes the proof.

$\Box$