Like a pyramid but not quite

Since $\cfrac{32}{56} = \cfrac{24 - x}{36 - x}$ solves to $x = 8$ , that means $IJ = 8$ , and we can take out a triangular prism cross-section with a lateral height of $8$ from $EFGH-IJ$ and combine the remaining sections to create a pyramid with a $32 \times 16$ rectangular base:

By proportions, the height $h$ of the pyramid fulfills $\cfrac{32}{56} = \cfrac{h}{h + 24}$ , which solves to $h = 32$ .

Therefore, the volume of $EFGH-IJ$ is $V = V_{\text{pyr}} + V_{\text{prism}} = \frac{1}{3} \cdot 32 \cdot 16 \cdot 32 + \frac{1}{2} \cdot 32 \cdot 32 \cdot 8 = \boxed{9557} + \frac{1}{3}$ .

Taking a horizontal cross section of polyhedron $EFGH-IJ$ at elevation $z$ , it will be a rectangle having dimensions $L(z) \times W(z)$

The length $L(z) = 56 + \dfrac{(32 -56)}{h} z = 56 - z$ , and the width $W(z) = 36 + \dfrac{24 - 36 }{h} z = 36 - \frac{1}{2} z$

Hence, the volume is given by $V = \displaystyle \int_{h}^{56} A(z) dz = \int_{24}^{56} (56 - z)(36 - \frac{1}{2} z ) dz$

The integral is straightforward to perform and yields, $V = 9557 \frac{1}{3}$ . Thus, $N = \boxed{9557}$