jmc

Let $h$ be the number of 4 unit line segments and $v$ be the number of 5 unit line segments. Then $4h+5v=2007$. Each pair of adjacent 4 unit line segments and each pair of adjacent 5 unit line segments determine one basic rectangle. Thus the number of basic rectangles determined is $B=(h-1)(v-1)$. To simplify the work, make the substitutions $x=h-1$ and $y=v-1$. The problem is now to maximize $B=xy$ subject to $4x+5y=1998$, where $x$, $y$ are integers. Solve the second equation for $y$ to obtain $$y=\frac{1998}{5}-\frac{4}{5}x,$$and substitute into $B=xy$ to obtain $$B=x\left(\frac{1998}{5}-\frac{4}{5}x\right).$$The graph of this equation is a parabola with $x$ intercepts 0 and 999/2. The vertex of the parabola is halfway between the intercepts, at $x=999/4$. This is the point at which $B$ assumes its maximum.

However, this corresponds to a nonintegral value of $x$ (and hence $h$). From $4x+5y=1998$ both $x$ and $y$ are integers if and only if $x\equiv 2(\mathrm{mod}5)$. The nearest such integer to $999/4=249.75$ is $x=252$. Then $y=198$, and this gives the maximal value for $B$ for which both $x$ and $y$ are integers. This maximal value for $B$ is $252\cdot 198=\boxed{49896}.$