The point (a,b) is not visible if and only if a and b are not coprime. The square with vertices (a,b) ,(a+n,b),(a+n,b+n),(a,b+n) satisfies the condition of the problem if for each i and j a+i and b+j are not coprime but by chinese remainder theorem if we pick $p_{ij}$ to be different primes we can set $p_{ij}$ to divide a+i and b+j for all i,j=0,1,2...n. So no points in this square is visible from the origin

How does the above construction ensure that the visible points when scaled down will also be contained inside a square with area $n^2$?

The key is to see that if $\gcd(x_i,y_i)\ne 1$ then a point is invisible. Then, we know by CRT there exists a solution with primes $p_l\forall 1\le l\le n^2$ s.t. if $x_1,y_1$ is the coordinate of the bottom left corner, we have $$x_1+i\equiv0\pmod{\prod_{k=1}^{n}p_{k+in}},y_1+i\equiv0\pmod{\prod_{j=0}^{n-1}p_{i+1+jn}}\forall 0\le i\le n-1.$$

A Nice Problem : Note that, a point is visible if $\gcd(x, y)=1$. So, we must have a common factor to both $x, y$ co-ordinates inorder for it to be invisible. Let $(x_1, y_1)$ denote the left-bottom most point of the square. Notice that, due to Chinese Remainder Theorem, we can find primes $p_{i, j}$ for $1 \le i, j \le n$ such that: $$x_1 \equiv 0 \pmod {p_{1, 1} p_{1, 2} \cdots p_{1, n}}$$$$x_1+1 \equiv 0 \pmod {p_{2, 1} p_{2, 2} \cdots p_{2, n}}$$$$\cdots$$$$x_1+n-1 \equiv 0 \pmod {p_{n, 1} p_{n, 2} \cdots p_{n, n}}$$$$y_1 \equiv 0 \pmod {p_{1, 1} p_{2, 1} \cdots p_{n, 1}}$$$$y_1+1 \equiv 0 \pmod {p_{1, 2} p_{2, 2} \cdots p_{n, 2}}$$$$\cdots$$$$y_1+n-1 \equiv 0 \pmod {p_{1, n} p_{2, n} \cdots p_{n, n}}$$ Hence we can find such an invisible square and we are done.

CRT FORCE FTW!!!! Consider pairwise distinct primes $\{p_{i,j}\}_{1 \le i, j \le n+1}$. Choose integers $x$ and $y$ such that $x \equiv -i+1 \pmod{p_{i,j}}, y \equiv -j+1 \pmod{p_{i,j}}$ for all $(i, j) \in \{1, 2, \dots, n+1\}^2$. Then the square with diagonally opposite vertices $(x, y), (x+n, y+n)$ works.