Inspired by Marte Reece 4.0

Since $r>1$ , we can suppose that $N\geq7$ . Let's number the grid points in the clockwise order. Let $P_1, P_2, P_3, \cdots, P_{N}$ be the points. Now consider the $P_1P_3, P_2P_4, P_3P_5, P_4P_6, \cdots, P_nP_2$ arces. These arces covers the cirlce's arc two times, so there will be an arc (let assume that $P_1P_3$ ), which length is at most $\dfrac{4\pi r}{N}$ . The value of $[\triangle P_1P_2P_3]$ is the maximum when $P_2$ is the midpoint of the $P_1P_3$ arc.

Since in a circle with radii $r$ , a convex $\alpha$ 's center angle's chord's is at most $2r\sin\alpha/2$ , $[P_1P_2P_3]=\dfrac{abc}{4r}=\dfrac{2r\sin\frac{\pi}{N}*2r\sin\frac{\pi}{N}*2r\sin\frac{2\pi}{N}}{4r}$

Since if $0\leq\alpha\leq\dfrac{\pi}{2}$ , then $\sin\alpha\leq\alpha$ , $[P_1P_2P_3]\leq\dfrac{2r\sin\frac{\pi}{N}*2r\sin\frac{\pi}{N}*2r\sin\frac{2\pi}{N}}{4r}=\dfrac{4r^2\pi^3}{N^3}$

Since a grid triangle's area is minimum half, $\dfrac{1}{2}\leq\dfrac{4r^2\pi^3}{N^3}\Leftrightarrow \large N\leq 2\pi\sqrt[3]{r^2}$

Therefore the statement is true.