jmc

There are $223\cdot 9=2007$ squares in total formed by the rectangle with edges on the x and y axes and with vertices on the intercepts of the equation, since the intercepts of the lines are $(223,0),(0,9)$. Count the number of squares that the diagonal of the rectangle passes through. Since the two diagonals of a rectangle are congruent, we can consider instead the diagonal $y=\frac{223}{9}x$. This passes through 8 horizontal lines ($y=1\dots 8$) and 222 vertical lines ($x=1\dots 222$). At every time we cross a line, we enter a new square. Since 9 and 223 are relatively prime, we don’t have to worry about crossing an intersection of a horizontal and vertical line at one time. We must also account for the first square. This means that it passes through $222+8+1=231$ squares. The number of non-diagonal squares is $2007-231=1776$. Divide this in 2 to get the number of squares in one of the triangles, with the answer being $\frac{1776}{2}=\boxed{888}$.