But they must be integers

By Pick's theorem the area of a simple polygon with vertices as lattice points in the plane is $A = i + \frac{b}{2} - 1$ where $i$ is the number of interior lattice points and $b$ is the number of boundary lattice points. For a quadrilateral with integer coordinates this at least $0 + \frac{4}{2} - 1 = \boxed{1}$ with equality occurring for any example with no interior lattice points and four boundary lattice points (e.g. a parallelogram with vertices at $(0,0), (1,0), (a,1), (a+1,1)$ where $a$ is an integer).

A quadrilateral has four vertices. Therefore possible vertices would be ( a, a), ( a+1, a), ( a, a+1) and ( a+1, a+1), where a is an integer. Area of this quadrilateral(square) is 1 square area. So, answer is 1.