Their sum is 46. Therefore, for any total area smaller than 46, the professor's statement must necessarily hold.
The smallest possible area that's at least 46 is 50. Does the professor's statement hold for N=10?
No, it does not. A 5x10 rectangle can be split up into 1x1, 1x2, 1x3, 1x4, 1x5, 2x2, 2x3, 2x4, 1x7, 1x10 rectangles, none of which are congruent.
For a 5x9 rectangle, it can be split up into 9 rectangles of sizes 1x1, 1x2, 1x3, 1x5, 1x7, 1x9, 2x2, 2x3, 2x4. This allows one single rectangle to be added to produce 5xN for any N>10 (but not N=10 itself, since it would require two 5x1 rectangles). Picture for N=11:
So the professor's statement doesn't hold for N=10, nor for any N>10. That makes the answer 9.