jmc

Let the two points $P$ and $Q$ be defined with coordinates; $P=(x_1,y_1)$ and $Q=(x_2,y_2)$ We can calculate the area of the parallelogram with the determinant of the matrix of the coordinates of the two points(shoelace theorem). \det \left($$\begin{array}{c}P\\ Q\end{array}$$\right)=\det \left($$\begin{array}{cc}{x}_1& y_1\\ x_2& y_2\end{array}$$\right). Since the triangle has half the area of the parallelogram, we just need the determinant to be even. The determinant is $$(x_1)(y_2)-(x_2)(y_1)=(x_1)(2009-41(x_2))-(x_2)(2009-41(x_1))=2009(x_1)-41(x_1)(x_2)-2009(x_2)+41(x_1)(x_2)=2009((x_1)-(x_2))$$ Since $2009$ is not even, $((x_1)-(x_2))$ must be even, thus the two $x$'s must be of the same parity. Also note that the maximum value for $x$ is $49$ and the minimum is $0$. There are $25$ even and $25$ odd numbers available for use as coordinates and thus there are ${(}_{25}{C}_2)+{(}_{25}{C}_2)=\boxed{600}$ such triangles.