Consider the below coloring , In both of (1) and (2) we have an even number of colored cells but not in the (3) . On the other hand , number of all the colored cells is even .So we get desire result . (Because if the sum of some integers being even then there are an even number of odd numbers among them.)

Attachments:

mahanmath wrote: Consider the below coloring , In both of (1) and (2) we have an even number of colored cells but not in the (3) . On the other hand , number of all the colored cells is even .So we get desire result . (Because if the sum of some integers being even then there are an even number of odd numbers among them.) Nice! You can even simplify the coloring:

Attachments:

Give each square a coordinate $(a,b)$. Color each square white, gray, or black as follows. If $(a,b)$ is tiled with a square: gray If $a+b$ is odd and the line through it is of positive slope: white If $a+b$ is odd and the line through it is of negative slope: black If $a+b$ is even and the line through it is of positive slope: black If $a+b$ is even and the line though it is of negative slope: white Define two squares to be "connected" if there is a tile that lies on both squares with the property that connected tiles are of opposite colors if and only if they share a parallelogram tile. Each square is connected to either zero or two other squares. Thus the rectangle is the union of disjoint cycles. Each cycle has an even number of pairs of connected squares of different colors. Hence, each cycle has an even number of parallelograms. Thus, in the whole rectangle there are an even number of parallelograms, as desired.

paint the grid points black and white in the normal way,and it suffices to consider the oblique dissection line segments with two white points. since there are an even number of them(this is because every of these segments belongs to two subfigures,and none of them appear on the boundary),and the configuration of (1) and (2) both have an even number of them,(3) has an odd number of them.so the number of (3) is even.

we can also get the number of $(1)$ is also even. consider a single square $ABCD$,if it is divided by diagonal $AC$,we write number 1 in these two isosceles triangles,if it is divided by diagonal $BD$,we write number 0 in these two triangles,otherwise the number in these two trianglesare arbitrary (but must be same!),so the sum of number in (1) is odd,and the sum of two numbers in $(2),(3)$ is even,so the number of $(1)$ is even

I think my solution is different from the ones above, so here it goes: Consider the midpoints of the diagonals of length $\sqrt{2}$ drawn as the rectangle is divided into the described figures. Now we connect with a red segment: 1) The midpoints of the two equal sides of length $\sqrt{2}$ in any triangle from the partitioning. 2) The midpoints of the two opposite sides of length $\sqrt{2}$ of any parallelogram in the partitioning. Note that each of these segments is of length 1. Also note that every diagonal drawn is the side of exactly two figures of the above type, so what we get is one or more closed red lines, the union of which goes through every midpoint of a diagonal drawn exactly once. Furthermore, these closed lines are disjoint (since every midpoint of a diagonal is an endpoint of exactly two red segments) and each of these closed lines is a polygon with angles only $90^{\circ}$ and $270^{\circ}$. Since every segment it's constructed of is of length 1, it follows that each of the closed red lines is of even length. Furthermore, every time such a line crosses a triangle, the orientation of the diagonal it crosses changes and when the red line crosses through a parallelogram, this orientation remains the same. Hence because all the lines are closed and the orientation of of the diagonal in the end should be the same as in the beginning, each red line crosses an even number of triangles. Since its total length is even, it must cross an even number of parallelograms. Because every parallelogram is crossed by exactly one red line, we are done. $\square$