Solution:

Use similar triangles as shown in the diagram. The left side of the large square has length $4x+2y$, and the bottom side has length $3y+x$. Since these are equal, $3y+x=4x+2y$, so $y=3x$. The side length of the large square is therefore $10x$.

In the upper part of the figure, let the marked horizontal segment be $m$. The two right triangles formed by the sides of rectangle $R$ have parallel corresponding sides and equal hypotenuses, so they are congruent. This gives the lengths shown below.

Similar triangles give $$\frac{m}{3x}=\frac{4x-\frac m3}{6x-m}.$$ Hence $6xm-m^2=12x^2-xm$, so $$m^2-7xm+12x^2=0=(m-3x)(m-4x).$$

If $m=3x$, rectangle $R$ has side length $3\sqrt2x$ in both directions, so its area is $18x^2$. If $m=4x$, its side lengths are $5x$ and $\frac{10}{3}x$, so its area is $\frac{50}{3}x^2$.

The two possible areas sum to $\frac{104}{3}x^2$. Since the $1\times3$ rectangle gives $x^2+(3x)^2=1$, we have $x^2=\frac1{10}$. The sum of the possible areas is $$\frac{104}{3}\cdot\frac1{10}=\frac{52}{15}.$$

Thus $m+n=52+15=67$, and the answer is E .