A variant of Conway's Game of Life: any cell with more than 3 live neighbours becomes a live cell and no live cell dies. How to make more cells live? In Conway's Game of Life, we got an infinite board (two-dimensional orthogonal grid of squares).
Every cell in the board interacts with its eight neighbors, which are the cells that are horizontally, vertically, or diagonally adjacent.
We first make $n$ many cells live following some pattern.
At each step in time, any (dead) cell with no less than four live neighbors becomes a live cell, and a live cell never dies.
It is easy to see the number of live cells no more increased for a sufficiently long time.
How do give an initial pattern, such that there will be the maximum number of live cells finally?
For example, let $n = 4$ and the initial pattern is as follows (live cells are green.)

The second pattern becomes

and then

Finally, we get the following pattern and it never changes.

 A: The optimal number of cells is $\frac47n^2(1+o(1))$. Sorry, in the lower bound I do not care about $o(n^2)$; perhaps the estimate can be done more precise.
We always assume that the new live cells appear one by one (this does not affect the final state).
1. Call a pair of neighbours diverse if they have different colors. Let $D$ denote the current number of diverse pairs. Then, as a new live cell appears, the value of $D$ does not increase (at least 4 pairs stop being diverse, and at most 4 new ones appear). So this total number does not increase.
Initially, we have $D\leq 8n$. So we are to estimate the number of cells in the final arrangement $A$ containing at most $8n$ diverse pairs.
2. Suppose that there exists an empty diagonal (say, in the upper-right direction) separating $A$ into two nonempty parts. Then one can shift the part under the diagonal by one left (both in the initial and in the final arramgements). It is easy to see that the new final arrangement is also reachable from the new initial one (but it may become not final, thus increasing the final number of cells). Performing such operations, we eventually arrive at an arrangement with no empty diagonals. Similarly, we get rid of empty cseparating columns and rows.
3. Now assume that $A$ fits into an $a\times b$ rectangle with $a\leq b$; then there are already $2(a+b)$ vertical and diagonal diverse pairs, two per nonempty row or column. There are several empty diagonals in this rectangle (say, $d_1$ in one direction and $d_2$ in the other); since no such diagonal is separating, they are situated at the corners. Then the total number of diagonal diverse pairs is at least $2(a+b-1-d_1)+2(a+b-1-d_2)$. So the total number of diverse pairs is at least
$$
  2(a+b)+2(a+b-1-d_1)+2(a+b-1-d_2)=2(3s-d-2)\leq 8n,
$$
where $s=a+b$ and $d=d_1+d_2$. So
$$
  3s-(d+2)\leq 4n.
$$
Assyme that there are $c_i$ empty diagonals at the $u$th corner of the rectangle (the numeration is cyclic), so that $c_1+c_2+c_3+c_4=d$. Notice that each boundary row/column of the rectangle contains at least one cell, so $c_1+c_2,c_3+c_4\leq a-1$. Therefore, $d\leq 2(a-1)\leq s-2$.
Then the total number of cells in $A$ is at most
$$
  N\leq ab-\sum_{i=1}^4\frac{c_i(c_i+1)}2
  \leq \frac{s^2}4-\frac{d^2}8-\frac{d}2=\frac{2s^2-(d+2)^2+4}8.
$$
Recall that $d+2\geq 3s-4n$. If $3s\leq 4n$, then we already have $N\leq s^2/4\leq 4n^2/9$. Otherwise, we get
$$
  N\leq \frac{2s^2-(3s-4n)^2+4}8
  =\frac{-7\left(s-\frac{12}7n\right)^2+\frac{32}7n^2+4}8
 \leq \frac47n^2+\frac12,
$$
which establishes the upper bound.
4. To achieve this bound, we should get $s\approx\frac{12n}7n$ (moreover, $a\approx b\approx \frac67n$) and $d+2\approx\frac87n$ (moreover, all $c_i\approx \frac27n$), so the final arrangelemt should look like an octagon whose sides are (almost) vectors of the form $(x,y)$, where $x,y\in\left\{\pm\frac27n,0\right\}$. Such octagon can be reached inductively, as shown below.

A: There is a simple construction leading to a lower bound $\frac{1}{2}n^{2} - O(1)$.
The dark green cells forms the initial pattern, and all the green cells form the final pattern.
Formally, let the initial pattern be $I_{m} = \{(0, \pm 2k) : k = 0,1,\ldots,m \} \cup \{(\pm (2k + 1), 0) : k = 0,1, \ldots, m \}$.
Thus $n = |I_{m}| = 4m - 1$ and the number of the cells in the final pattern is
$$8m^{2} - 4m - 1 = 8 \cdot \left( \frac{n + 1}{4} \right)^{2} - 4 \cdot \left( \frac{n + 1}{4} \right) - 1 = \frac{1}{2} n^{2} - \frac{3}{2}$$.


Below is an improving approach leading to a lower bound $\frac{4}{7}n^{2} - O(n)$.
Consider an octagonal $X(\alpha)$ on the board, where $\alpha = (a_{1}, b_{1}, a_{2}, b_{2}, a_{3}, b_{3}, a_{4}, b_{4})$ is the length of the sides of $X(\alpha)$ with clockwise order and $a_{1}$ is the length of the top side.
Specifically, $a_{i}$ is the length of the sides that are parallel or perpendicular to the grid lines, and $b_{i}$ is the length of the sides that are parallel or perpendicular to the grid diagonals.
Let $\mathbf{e}_{1} = (1,0,1,0,1,0,1,0)$ and $\mathbf{e}_{2} = (0,1,0,1,0,1,0,1)$.
Denote $X(a, b) = X(a\mathbf{e}_{1} + b\mathbf{e}_{2})$ be an octagonal such that the length of the side is $(a, b, a, b, a, b, a, b)$.
Specially, if $a = 0$, $X(a, b)$ is a diamond; if $b = 0$, $X(a, b)$ is almost a square; if $a = b$, $X(a, b)$ is a regular octagonal.
Suppose that $a = k_{1}n + O(1)$ and $b = k_{2}n + O(1)$, the number of the cells in $X(a, b)$ is $|X(a, b)| = a^{2} + 2b^{2} + 4ab - O(a + b)$.
Suppose that for a certain initial pattern, the final pattern is exactly $X(a, b)$.
If we can make more cells live in the initial pattern, then the final pattern becomes larger.
We now define two types of operations $f(\cdot)$ and $g(\cdot)$ on $X(a, b)$ such that if there are four more live cells in the initial pattern, then the final pattern become $X(f(a, b))$ and $X(g(a, b))$, respectively.
Type I operation: The grey cells form an $X(a, b)$.
Make every dark green cell a live one.
Finally, every light green cell will become a live one.
After doing this operation
$$f \colon (a, b) \mapsto (a + 4, b - 1)$$

Type II operation: The grey cells form an $X(a, b)$.
Make every dark green cell a live one.
Finally, every light green cell will become a live one.
After doing this operation
$$g \colon (a, b) \mapsto (a, b + 2)$$

Define $F = f \circ f \circ g$ and $G = g \circ g \circ g \circ g$.
We know that $F(a, b) = (a + 8, b)$ and $G(a, b) = (a, b + 8)$.
Suppose the final pattern is $X_{a, b}$.
Hence, we can make 12 or 16 more cells live in the initial pattern, then the final pattern becomes $X(a, b + 8)$ or $X(a + 8, b)$, respectively.
For small constants $a_{0}, b_{0} > 0$ and $k_{1}, k_{2} \in [0, 1]$, let
$$n = \left(\frac{12}{8}k_{1} + \frac{16}{8}k_{2}\right)m + \left|X(a_{0}, b_{0})\right| = \left(\frac{3}{2}k_{1} + 2k_{2}\right)m + \left|X(a_{0}, b_{0})\right| $$
We can construct a final pattern which is exactly $X(k_{1}m + a_{0}, k_{2}m + b_{0})$.
Therefore, we have that the number of live cells in the final pattern is
$$
\begin{aligned}
|X(k_{1}m + a_{0}, k_{2}m + b_{0})|
= &(k_{1}^{2} + 2k^{2}_{2} + 4k_{1}k_{2})m^{2} + O(m) \\
= &(k_{1}^{2} + 2k^{2}_{2} + 4k_{1}k_{2}) \left(\frac{n - O(1)}{\frac{3}{2}k_{1} + 2k_{2}}\right)^{2} + O(n) \\
= &\frac{k_{1}^{2} + 2k^{2}_{2} + 4k_{1}k_{2}}{\left(\frac{3}{2}k_{1} + 2k_{2}\right)^{2}}n^{2} - O(n) \\
\end{aligned}
$$
When $k_{1} = k_{2}$, it achieves the maximum value $\frac{4}{7}n^{2} - O(n)$.
