Prove that using lines parallel to the coordinate axes to divide the plane into some $2 \times 2$ squares, and then translating all the squares that cover the points of $K$ to the same square, since the area of $K$ is greater than 4, there must be a point in this square that is covered by the squares containing points of $K$ twice, i.e., there are two points $A \neq B$ in $K$, whose differences in both the x-coordinates and y-coordinates are multiples of 2. Since the figure $K$ is symmetric about the origin, the point symmetric to $A$ about the origin is $A^{\prime}$, then $A^{\prime} \in K$. Let $B\left(x_{1}, y_{1}\right), A\left(x_{2}, y_{2}\right)$, then $x_{1}-x_{2}=2 m, y_{1}-y_{2}=2 n, m, n \in \mathbf{Z}$. Therefore, $A^{\prime}$ is $\left(-x_{2},-y_{2}\right)$, so the coordinates of the midpoint of $A^{\prime} B^{\prime}$ are $\left(\frac{x_{1}-x_{2}}{2}, \frac{y_{1}-y_{2}}{2}\right)=(m, n) \neq(0,0)$. Since $K$ is a convex figure, $(m, n)$ is inside $K$.