If there exists some $c$-monochromatic horizontal row $y=k$, then if rows $y=k+1$ and $y=k+2$ are $\overline{c}$-monochromatic we can find $\overline{c}$-monochromatic triangles of any positive integer area, otherwise they must contain at least a $c$-point, and we can find $c$-monochromatic triangles of any positive integer area. So assume there exists no monochromatic horizontal row. Consider the horizontal row $y=0$. If it contains two $c$-points at distance $1$, then for any positive integer $n$, together with a $c$-point on row $y=2n$ they will make a $c$-monochromatic triangle of area $n$. Otherwise it will contain two $\overline{c}$-points at distance $2$, which for any positive integer $n$, together with a $\overline{c}$-point on row $y=n$ will make a $\overline{c}$-monochromatic triangle of area $n$.

An easy problem, but nice for a problem $1$. Suppose the opposite. Case $1$. We have $2$ points of the same color such that their coordinates are $(x,y)$ and $(x,y+1)$ or $(x,y)$ and $(x+1,y)$. WLOG, let them be $(x,y)$ and $(x+1,y)$, and let them be white. Now, there must exist a horizontal line with all blue points (if not that is a contradiction). So we have a blue horizontal line, and that means that all points that are not on that line are white, so contradiction. Case $2$. The points are white-blue after a chessboard fashion, and we can also have all triangles with one color, so done.

junioragd wrote: WLOG, let them be $(x,y)$ and $(x+1,y)$, and let them be white. Now, the horizontal line above the line of these two points must be blue I don't see how this follows. What if there's a white point on the line? You can still have many white points on the line (at least anything finite works) and you still won't make all necessary white triangles.

It's easy problem!!! We have a vertical line $x=0$ and this line has two points such that both are in same colour (for example, white) and distance between of the two points equal to $n$ or $2n$. (1) suppose that distance beetwen two points is equals to $2n$. Then all points on the lines $x=1$ and $x=-1$ are blue and we are done! (2) suppose that distance beetwen two white points equals to $n$. Then all points on the lines $x+2=0$ and $x-2$ are blue and we are done!

Is it two colors the maximum ? what is the minimum number of colors such there is a configuration with no monochromatic triangle for every $n$ ?

The maximum is two colors,here is a counterexample for $3$ colors 123123... 231231... 312312... Now,in this pattern we don't have a triangle of area $1$.

mathuz conmitted the same error as junioragd.

Please help me check whether this is true or not. Suppose there exist 2 adjacent points with same colour, WLOG let them be white and belong to a horizontal line y=k. If all points on the line y=k+1 are white then choose white, else there must be a blue point on line y=k+1. Then if all points on the line y=k+2 is blue, then choose blue, else there must be a white point on y=k+2. Repeat to deduce that there is a white point on lines y=k+2m for all m. Then we choose the 2 adjacent white points, and any white point with height 2m, then the area is m for all positive integers m, which settle this case. Else, no two adjacent points have same colour (this applies to both horizontal and vertical.), e.g: the colouring is chessboard style colouring, which clearly possible to choose monochromatic triangles with any integral area and we are done. QED

Assume the contrary. Take two adjacent points, WLOG they are both blue and at $(x,0), (x+1,0)$ (if none exist, the colors are in a checkerboard pattern and the result immediately follows). Then, there exists some line $y = 2k$ that is all white (or else $(x, 0), (x+1, 0), (a, 2n)$ make a triangle of area $n$ for all $n$), which means that $y = 2k-1$ and $y = 2k+1$ must be all blue (or else we can take $(b, 2k), (b, 2k\pm1), (b+2n, 2k)$ to make a triangle of area $n$ for all $n$). But this means we can take $(b, 2k-1), (b, 2k+1), (b+n, 2k-1)$ to make a triangle of area $n$ for all $n$, a contradiction. Thus, the problem statement must be true.

Notice that, if for every neighbouring points are of different color, one can simply select a line $\ell$, and two adjacent points of the same color on $\ell$, and then, another point on the line of distance $2n$ from $\ell$, of the same color. Clearly, such an assignment works. Hence, assume that there is an adjacent, monochromatic pair, say colored with blue. Assume that, this process fails, namely, for some $n$, we are not able to find such a point. Now, focus on the line of distance, $\ell_{2n}=2n$. Clearly, all the points on this line are red (otherwise, we are good to continue). Now, look at the line, $\ell_{2n-1}$, which is of distance $1$ from $\ell_{2n}$, and $2n-1$, from $\ell$. If there is a red point on this line, we simply conclude, since we can instead consider red triangles, whose one vertex is the red point on $\ell_{2n-1}$, and by modifying the base via selecting two vertices on $\ell_{2n}$, we always find a monochromatic triangle (red) with area $n$ for any positive integer $n$. Thus, all points of $\ell_{2n-1}$ are blue. Similarly, consider all points on $\ell_{2n-2}$. If there is a blue point, same discussion applies, and we stop. Hence, $\ell_{2n-2}$ is all red. Continuing this process, we arrive at, $\ell_1$, which is all blue. Now, one of the blue points on $\ell$, and modulating the base on $\ell_1$, we can construct all such desired triangles.

$2x2n$ need to look at the shape