Solution:
Clearly, $n=4$ is too small, for the 4 points could be vertices of a unit square, and all the possible triangles will have areas of $1/2$. We will show that $n=5$ works. Every lattice point falls into one of four parity classes: (even, even), (even, odd), (odd, even), and (odd, odd). For example the point $(9,8)$ is in the (odd, even) class. By the pigeonhole principle, among the 5 lattice points there will be at least two points, $A$ and $B$, in the same parity class. We claim that if $C$ is any other lattice point, the triangle $A B C$ will have integral area.

We will use the following simple lemma:
If the vertices of a triangle are all lattice points, and one side is horizontal or vertical with even length, then the triangle must have integer area.
To see why the lemma is true, let $P Q R$ be the triangle with $P Q$ even, and, say, horizontal. Let $h$ be the length of the altitude from $R$ to $P Q$. Clearly, $h$ is an integer, since it is equal to the absolute value of the difference in $y$-coordinates of $R$ and $P$. Then $[P Q R] = P Q \cdot h / 2$, and the lemma follows. (We are using the notation $[P Q R]$ to denote the area of triangle $P Q R$.)

Now we will show that $[A B C]$ must be an integer. If $A B$ is horizontal or vertical, then the length of $A B$ is even (since the endpoints are in the same parity class), so the lemma immediately implies that $[A B C]$ is an integer.
If $A B$ is neither horizontal nor vertical, choose the lattice point $D$ such that $A D B$ is a right triangle with right angle at $D$ (there may be more than one choice).

Then
$$[A B C] = [A C B D] - [A D B].$$

Since $A D$ and $D B$ both have even length, $[A B C]$ is an integer. Furthermore,
$$[A C B D] = [C A D] + [C D B],$$
and by the same reasoning $[C A D]$ and $[C D B]$ are both integers.