Solution:

Suppose an $m \times n$ rectangle could be tiled as described. We will establish a contradiction by counting gnomon vertices.

A gnomon cannot touch a side of the rectangle along a length $1$, because then the gnomon that fitted under the overhang would form a rectangle with the first. So each gnomon along a side of the rectangle touches it along a length $2$. So $m$ and $n$ must be even. Put $m = 2M$, $n = 2N$. There are $(2M - 1)(2N - 1)$ gridpoints inside the rectangle. None of these points can have $4$ gnomon vertices. But it is easy to see that they cannot have $3$, because the angle inside a gnomon at a vertex is either $90^{\circ}$ or $270^{\circ}$. So they have at most $2$ gnomon vertices each, or $2(2M - 1)(2N - 1)$ in total. There is only one gnomon at each of the $4$ corners, or $4$ gnomon vertices in total. Along the sides there are alternately $2$ and $0$, so at most $4(M - 1) + 4(N - 1)$ in total, giving a grand total of $\leq 2(2M - 1)(2N - 1) + 4(M - 1) + 4(N - 1) + 4 = 8MN - 2 < 8MN$.

On the other hand, there are $4MN / 3$ gnomons each with $6$ vertices, a total of $8MN$. Contradiction.