We will construct our set $\{A_n=(x_n,y_n)\}$ as a set of lattice points. the area of every triangle will automatically be rational (half an integer, actually), so all we need to do is ensure the fact that no three points are colinear and the distance between any two of them is irrational. The base step $n=3$ is easy: take $A_1=(0,0),\ A_2=(1,1),\ A_3=(3,2)$. Now suppose $n\ge 4$ and we have constructed the lattice points $A_1,A_2,\ldots,A_{n-1}$. For $i=\overline{1,n-1}$ choose primes $p_i\equiv 1\pmod 4$, and choose $a_i,b_i$ s.t. $p_i\not|a_i,b_i,\ p_i|a_i^2+b_i^2$. After that, we can find $k_i$ s.t. $p_i\|(a_i+k_ip_i)^2+b_i^2$ (i.e. $p_i$ divides $(a_i+k_ip_i)^2+b_i^2$, but $p_i^2$ does not). Finally, by the Chinese Remainder Theorem, we can find $x_n,y_n$ s.t. $x_n-x_i\equiv a_i+k_ip_i\pmod{p_i^2},\ y_n-y_i\equiv b_i\pmod{p_i^2},\ \forall i\in\overline{1,n-1}$. The numbers $(x_n-x_i)^2+(y_n-y_i)^2$ will be divisible by $p_i$ but not $p_i^2$, meaning that all the distances $d(A_n,A_i),\ i=\overline{1,n-1}$ are irrational. This process may yield colliniarities $A_n,A_i,A_j,\ i\ne j\in\overline{1,n-1}$, but this can easily be repaired. Take another prime $p$ which divides none of the differences $x_i-x_j,y_i-y_j,\ i\ne j\in\overline{1,n-1}$, different from $p_i,\ \forall i$ and we also throw in the conditions $p|x_n-x_i,\ p\not|y_n-y_i,\ \forall i\in\overline{1,n-1}$. This will ensure that $\frac{y_n-y_i}{x_n-x_i}\ne\frac{y_i-y_j}{x_i-x_j},\ \forall i\ne j\in\overline{1,n-1}$.

take $C$ an integer very superior to $n$. Then the points $A_1=(C+1,C^{C+1})$, $A_2=(C+2,C^{C+2})$,..,$A_n=(C+n,C^{C+n})$. Of course the triangles will not be degenarated. And if $C$ is large enough, $d(A_i,A_j)^2=(i-j)^2+(C^{C+i}-C^{C+j})^2$ is an integer but not a perfect square.

Take an equilateral triangle with area 1 and thus side length $ 2\sqrt {\frac {1}{\sqrt {3}}}$, and take any set of points with rational areal co-ordinates with respect to this triangle and no three collinear. I claim this set satisfies the desired properties. Proof: Rational areal co-ordinates, and the rationality of the area of our triangle of reference, imply that every triangle comprised of three points of such a set, being the determinant of a rational matrix multiplied by a rational scalar, is rational in area. The distance d between any two rational points (a,b,c),(p,q,r) (with rational displacement vector (a-p,b-q,c-r)) is given by: \[ d^2 = \frac {4}{\sqrt 3} ( - (a - p)(b - q) - (b - q)(c - r) - (c - r)(a - p)) \not \in \mathbb{Q} \]

I have another solution: just take the integer points of a parabola $ y = x^2$. The area is clearly rational (in fact, the area of lattice polygons is always rational), but the distances between them are not: $ (a,a^2),(b,b^2); (a - b)^2 + (a^2 - b^2)^2 = (a - b)^2(1 + (a + b)^2)$ is not a perfect square. I think I can do it for other polynomials, but it wasn't test

By shoelace, any triangle which has three lattice points as vertices must be rational. Thus, it is sufficient to find a set of $n$ lattice points which have pairwise irrational distance. For any lattice point $(x,y)$, we let $S_{(x,y)}$ denote the set of lattice points which have an irrational distance to the point $(x,y)$. Notice that $S_{(x,y)}$ is just $S_{(0,0)}$ shifted by $x$ and $y$. Claim: For any positive integer $N$, there exists an $N \times N$ lattice grid which is contained entirely in $S_{(0,0)}$ Clearly true since points $(a,b) \in \mathbb{Z}^2$ where $\sqrt{a^2+b^2}$ is an integer has density $0$. $\square$ Thus, we proceed with induction. Take a set of $k$ points $\{P_1,P_2, \ldots, P_k \}$ which can be enclosed by a square of sidelength $M$. Then, we can find the same squares in $S_{P_1}, S_{P_2}, \ldots S_{P_k}$ of sidelength $M+1$. All of these squares have at least common lattice, and thus has irrational distance to all points chosen in the set. $\blacksquare$

Solution from Twitch Solves ISL: Note that any triangle with vertices which are lattice points has rational area. So we'd be done if we could prove: Claim: Given a finite set $S$ of lattice points, we can add one lattice point $P$ such that $P$ is irrational distance from every point in $S$. Proof. Suppose $S$ lies inside $[1,n] \times [1,n]$. Let $a^2+b^2=c^2$ by a primitive Pythagorean triple with the additional property that $a/b \neq x/y$ for any $(x,y) \in S$; this is possible there are infinitely many primitive Pythagorean triples. Consider large $R$; I claim that $P = \left( aR, bR \right)$ is okay if $R$ is sufficiently large. The squared distance from $P$ to a point $(u,v)$ is \[ D = \left( aR-u \right)^2 + \left( bR-v \right)^2 = (cR)^2 - 2(au+bv)R + (u^2+v^2) \]So for this to be square, it must be inside the set \[ D \in \{(5R-(a+b)n)^2, (5R-(a+b)n+1)^2, \dots, (5R+2n)^2 \}. \]In other words, if $R$ is large enough, then $D$ is not a square unless $D$ is actually identically the square of a polynomial in $R$. For that to be the case, we would need \[ D = \left( cR - \sqrt{u^2+v^2} \right)^2 \]which would only happen if \[ c\sqrt{u^2+v^2} = au+bv \implies 0 = c^2(u^2+v^2) - (au+bv)^2 = (bu-av)^2 \]which doesn't happen because we assured $a/b \neq u/v$. Hence for each $(u,v) \in S$, there exists a constant $C_{u,v}$ for which $R > C_{u,v}$ guarantees $P$ is irrational distance from $(u,v)$. Take the max of these finitely many constants to finish. $\blacksquare$

Let $L_t=n^{n^{n^{n^{n^t}}}}.$ Then, suppose the points are $(1,L_1),(2,L_2)\cdots (n,L_n)$. The distance between the points with x-coordinates a and b is $$\sqrt{(a-b)^2+(L_a-L_b)^2}.$$Note that $$(a-b)^2<<<<<<<2(L_a-L_b)+1,$$so the thing inside the root is not a perfect square, hence the distance is irrational. By shoelace, all areas are multiples of 1/2, hence done. (alternatively bary bash with respect to an equilateral triangle trivializes the problem )

Pick $n$ arbitrary non-collinear lattice points (Shoelace implies rational area). If $D$ is the largest squared distance among these, if $P>D$ is prime, scaling everything by $\sqrt{P}$ makes all lengths irrational but areas rational. When your solution is shorter than the problem statement...

Place on coordinate plane; induct: $n \longrightarrow n+1$. Take $A_{n+1}=(10^{10^{ \mid \max \{x_i\} \mid +10}}+\max \{x_i\},\max \{y_i\}+1)$ over $1 \leq i \leq n$. Trivially works by Shoelace and Pythag.

Consider the space $\mathbb{R}^{2n}$ of of $n$-tuples of points in the plane. Now, for each nonnegative rational $a$, we are excluding all tuples with points $P_1$, $P_2$, and $P_3$ forming a triangle of area $a$; for fixed points and fixed $a$ observe that that is a set with one less dimension so it has measure $0$; unioning over all triples of points and all nonnegative rationals $a$, noting this is a finite union, observe that we are excluding a set of measure $0$ and thus there must be an element outside of the set we are excluding, and this is the $n$-tuple of points we want (we take its underlying set).

There are lots of ways to do this, but here's one that uses $\nu_2$. We will use lattice points only, so the area condition is automatically satisfied. Now, let $(x_1, y_1), \dots, (x_n, y_n)$ be the $n$ points, and let $d_i = x_{i+1} - x_i$ and $e_i = y_{i+1} - y_i$. Pick $(d_1, d_2, \dots, d_{n-1}) = (1, 1, \dots, 1)$ and $$(e_1, e_2, e_3, \dots, e_{n-1}) = (1, 2^n + 1, 2 \cdot 2^n+1, \dots, (n-2) \cdot 2^n + 1).$$Notice that for two points $(x_i, y_i)$ and $(x_j, y_j)$, it suffices to show that $$(d_{i+1}+d_{i+2}+\cdots+d_j)^2 + (e_{i+1}+e_{i+2} + \cdots + e_j)^2$$is never a perfect square. This is clear because both have $\nu_2$ equal to $\nu_2(j - i)$ as all the $d_i$ and $e_i$ are $1$ mod $2^n$. Hence we have mod $4$ issues upon factoring out the $2^r$ term, done.

VerySus Consider the points $(x,x^2)$ for $x = 1,2,\dots,n.$ If $1 \le r,s \le n,$ then $$\sqrt{(r-s)^2 + (r^2-s^2)^2} = (r-s) \sqrt{(r+s)^2 + 1}.$$The expression in the square root is never a perfect square except when $r + s = 0,$ which obviously can't happen. The area of a lattice triangle is always rational, so this set of points satisfies the problem condition.

Consider the points of the form $(x,x^2)$ for $x=1,2,\ldots n$, as they are lattice points the area of any triangles formed by this points it´s rational, the distance between any of this points it´s equal to \[\sqrt{(x-y)^2+(x^2-y^2)^2}=(x-y)\sqrt{(x+y)^2+1}\]which is never rational

The answer is yes. Define $y_x = 1434^{1434^{1434^{1434^x}}}$. Then, consider the points $(i, y_i)$ for $1 \leq i \leq 1987$. Because $y_i$ is an integer for all integers $i$, then every point is a lattice point and all areas are rational by shoelace (in fact, the denominator must be 2 if present). However, the distance between any two points, $\sqrt{(i-j)^2 + (y_i - y_j)^2}$ cannot be an integer since $0 < (i-j)^2 \ll 2(y_i-y_j)+1$, so we are done.

Consider any subset of lattice points on the graph of $y=x^2$ for $x \ge 0$. By Shoelace, the area of each triangle formed by choosing three points is rational. Additionally, the distance between any two points $(a,a^2)$ and $(b,b^2)$ on the graph is \[\sqrt{(a-b)^2+(a^2-b^2)^2} = (a-b) \sqrt{(a+b)^2+1},\] which is clearly irrational. $\blacksquare$

weird construction