Darn, I misread the problem, I'll try to repair this solution later.

What do you mean with this: $ S_{\triangle ABC}=S $?

It means that area of triangle $\triangle ABC$ is equal to $S$. My "solution" is correct for different problem. It finds value $S$ such that for each coloring there is a triangle with area at most $S$. But I don't know how to solve the original problem

Bump. bump!

Since I was asked for a solution, here is the official one. (EDIT: It's a translation from Bulgarian, so the responsibility for any typos, bad wording, etc. is mine.) Suppose such $S$ exists. Then it should be a natural or a half of a natural number. Consider the following two colorings of the given points: (1) We color $(x,y)$ with a color $i\,,\, 1\leq i\leq 2$ iff $x\equiv i \pmod{2}$ (2) We color $(x,y)$ with a color $i\,,\, 1\leq i\leq 3$ iff $x\equiv i \pmod{3}$ The coloring (1) shows $S\in \{1,2,3,\dots\}$, whereas the other coloring (2) yields $S\in \{3/2,3,9/2,\dots\}$. Together it gives $S\geq 3$. Now, it's enough to prove that for any coloring (with 3 colors) there exists a triangle with area $3$. Note that there exists $d\in\{1,2,3\}$ and $x,y\in \mathbb{Z}$ s.t. $A :=(x,y)$ and $B:=(x+d,y)$ are of same color. Indeed, it's enough to consider the points $(0,0), (1,0),(2,0),(3,0)$. Now, denote the line through $A,B$ as $m$ and let $\ell$ be a line parallel and at distance $6/d$ to $m$. Then, if all three colors appear on $\ell$ we are done, since we can find a triangle with area $6$. So, suppose $\ell$ contains only $2$ colors. We say a distance $a$ is compatible with a line $\ell$ if there exist two monochromatic points at distance $a$ on $\ell$ . 1) Suppose, there exists $a\in\{1, 2,3,6\}$ which is compatible with the both colors of $\ell$. Then a line $p\parallel \ell$ and at distance $a$ at $\ell$ should be monochromatic, otherwise we will get a desired triangle. 2) If such a distance does not exists, then it can be checked (I omit here the details) that there is a color on $\ell$ compatible to all $a\in\{2,3,6\}$ (or we can find a monochromatic triangle with area $3$). In both cases there exists a line $p$ and a color $c$, compatible with all $d\in\{2,3,6\}$. Consider lines $u_1,u_2,u_3$ parallel to $p$ and of distances $1,2,3$ resp. at $p$ and on the same side of $p$. Then each of these $3$ lines should be colored in only two colors. Consider the meet points of the lines $x=0,x=3, x=6$ and $u_1,u_2,u_3$. It can be easily seen there exists a monochromatic triangle of area $3$ and vertices among these $9$ points.