Let's count the number of disjoint triangles made in this way. For 1 point we have 4 triangles, and for every point we add 2 triangles. (regardless wether the point is on a line or inside a triangle) So for $1999$ points, we have $4+2.1998=4000$ disjoint triangles, and a total area of $400$, so by pigeonhole there is at least one with area below $\frac1{10}$.

See also this thread: http://www.mathlinks.ro/Forum/viewtopic.php?highlight=points&t=52202

can anybody explain why counting like peter is valid. and it's no need to use the 4 vertices right? and what about 3 points lie in a line? edit: I'm just get it from the link above. thanks very much //so dumb question..

Solution. By induction, we can conclude if there are $n$ points inside the triangle, the square can be triangulated into $2n+2$ triangles. So we can triangulate $S$ with $2*1999+2=4000$ triangles. The average area is $$\frac{\text{Area Of the Square}}{\text{Number of Triangles}}=\frac{400}{4000}=\frac{1}{10}$$