Brazilian Math Olympiad

Since ${\left(\frac{16}{2\sqrt{3}}\right)}^2=\frac{64}{3}=21+\frac{1}{3}$ lies between $4.5^2=20.25$ and $5^2$ and the altitude of the triangle is $\frac{2\sqrt{3}\cdot \sqrt{3}}{2}=3$, we can cover a square with sidelength $16$ with $2\cdot 5\cdot \lfloor \frac{16}{3}\rfloor =60$ equilateral triangles. Since $\lfloor \frac{1000}{60}\rfloor =16$, by the pigeon hole principle there is an equilateral triangle that covers at least $17$ points.