Because any three points lattice points form a triangle with integer area, we will select $ n$ lattice points with irrational distance between any of them. We can build up this set by induction. Base Case: $ n=3$. Choose $ (0,0),(1,1),(2,4)$. Inductive step: Assume we have $ n$ points $ (x_1,y_1),(x_2,y_2),...,(x_n,y_n)$ with each with irrational distances from each other. Now choose $ y$ distinct from all the $ y_i$, and let $ a=\max |y-y_i|$. Now choose $ x$ sufficiently large such that $ x-\max x_i>\frac{a^2-1}{2}$ (meaning $ 2(x-\max x_i)+1>a^2$) and $ (x,y)$ is not collinear with any two points chosen already (this is possible because the first $ n$ points only determine $ \binom{n}{2}$ lines). Now I claim that the distance from $ (x,y)$ to $ (x_i,y_i)$ for any $ i$ is irrational. $ (x-x_i)^2+(y-y_i)^2>(x-x_i)^2$. $ (x-x_i)^2+(y-y_i)^2\le (x-x_i)^2+a^2<(x-x_i)^2+2(x-\max x_i)+1<(x-x_i)^2+2(x-x_i)+1=(x-x_i+1)^2$. $ (x-x_i)^2+(y-y_i)^2$ lies between two squares, so it cannot be a square; therefore the distance between $ (x,y)$ and $ (x_i,y_i)$ is irrational, so this completes the induction and so we can build a set with all triangle having rational area and all distances being irrational.

Also in IMO1987/5 Let there be a square lattice such that the length of each interval is $ \sqrt{3}$. i.e. $ (x,y)=(a\sqrt{3},b\sqrt{3})$ where $ a,b$ are integers. Then choose any $ n$ non-collinear points in the lattice. By the area formula, we get the area as $ \frac{1}{2}\sqrt{3}^{2}\cdot m$, where $ m$ is an integer. So the area is rational. But if $ k$ is the distance between some two points, then $ k^{2}=3p^{2}+3q^{2}$, where $ p$ and $ q$ are integers. It is easy to see that $ k$ cannot be rational unless it is an integer. Suppose $ k$ is an integer: take $ mod3$, we get $ 3\mid k$, then take $ mod3$ again, we get $ 3\mid p$ and $ 3\mid q$ and we get infinite descent. Contradiction, so $ k$ must be irrational.