Among all triples of points, we choose a triple $A,B,C$ so that $ABC$ has maximal area $F$. Obviously $F \leq 1$. Draw parallels to the opposite sides through A, B, C. You get $A_1B_1C_1$ with area $4F \leq 4$. We will show that $A_1B_1C_1$ contains all $N$ points. Suppose there is a point $P$ outside $A_1B_1C_1$. Then $ABC$ and $P$ lie on different sides of at least one of the lines $A_1B_1, B_1C_1, C_1A_1$. Suppose they lie on different sides of $B_1C_1$. Then $BCP$ has a larger area than $ABC$. This contradicts the maximality assumption about $ABC$

Hmm, this again?