Solution:
Answer: $127$

First we claim that if the lattice points in $[1,3] \times [1,7]$ are colored either black or white, then there exists a rectangle whose faces are parallel to the $x$ and $y$ axes, whose vertices are all the same color (a.k.a. monochromatic). Indeed, in every row $y = i$, $1 \leq i \leq 7$, there are two lattice points with the same color. Note there are $3$ combinations of $2$ columns to choose from (for the two similarly-colored lattice points to be in), and $2$ colors to choose from. By the Pigeonhole Principle, in the $2 \cdot 3 + 1 = 7$ rows two rows must have a pair of similarly-colored lattice points in the same columns, i.e. there is a monochromatic rectangle.

This shows that in each cross section $z = i$, $1 \leq i \leq N$ there is a monochromatic rectangle. Next, note there are $\binom{3}{2} \binom{7}{2}$ possibilities for this rectangle ($\binom{3}{2}$ ways to choose the $2$ $x$-coordinates and $\binom{7}{2}$ ways to choose the $2$ $y$-coordinates), and $2$ possible colors. Thus if $N = 2 \binom{3}{2} \binom{7}{2} + 1 = 127$, then by the Pigeonhole Principle there are two values of $i$ such that the same-colored rectangle has the same $x$ and $y$ coordinates in the plane $z = i$, i.e. there is a monochromatic rectangular prism.

For $N = 126$ the assertion is not true. In each cross section $z = i$, we can color so that there is exactly $1$ monochromatic rectangle, and in the $126$ cross sections, have all $126$ possible monochromatic rectangles represented. To do this, in each cross section we color so that each row has exactly $2$ lattice points of the same color, and such that $6$ of the rows give all possible combinations of $2$ points having the same color. This way, there will be exactly $1$ monochromatic rectangle in each cross section; we can obviously vary it for the different cross sections.