Let $A B C D$ be a rectangle with $A B=m$ and $A D=n$ where $m, n$ are natural numbers such that $m \geq n \geq 2$. Suppose that inside of the rectangle $A B C D$ is placed a rectangular lattice consisting of some identical squares whose areas are equal to 3, where $k$ of them are placed along the side $A B$ and $l$ of them along the side $A D$.

The sum of areas of all of these squares is equal to $3 k l$. Besides the obvious conditions $k \sqrt{3} \leq m$ and $(l+1) \sqrt{3} > n$ (2).

The proposed problem is to determine all pairs $(m, n) \in \mathbb{N} \times \mathbb{N}$ with $m \geq n \geq 2$, for which the ratio $R_{m, n}=\frac{3 k l}{m n}$ is equal to 0.5 where $k, l$ are natural numbers determined by the conditions (1) and (2).

Observe that for $n \geq 6$, using (2), we get $R_{m, n}=\frac{k \sqrt{3} \cdot l \sqrt{3}}{m n}>\frac{(m-\sqrt{3})(n-\sqrt{3})}{m n}=$ $\left(1-\frac{\sqrt{3}}{m}\right)\left(1-\frac{\sqrt{3}}{n}\right) \geq\left(1-\frac{\sqrt{3}}{6}\right)^{2}=\frac{1}{2}+\frac{7}{12}-\frac{\sqrt{3}}{3}>\frac{1}{2}+\frac{\sqrt{48}}{12}-\frac{\sqrt{3}}{3}=0.5$

So, the condition $R_{m, n}=0.5$ yields $n \leq 5$ or $n \in\{2,3,4,5\}$. We have 4 possible cases:

Case 1: $n=2$. Then $l=1$ and thus as above we get $R_{m, 2}=\frac{3 k}{2 m}>\frac{\sqrt{3} \cdot(m-\sqrt{3})}{2 m}=$ $\frac{\sqrt{3}}{2} \cdot\left(1-\frac{\sqrt{3}}{m}\right)$, which is greater than 0.5 for each $m>\frac{\sqrt{27}+3}{2}>\frac{5+3}{2}=4$, hence $m \in\{2,3,4\}$. Direct calculations give $R_{2,2}=R_{2,4}=0.75$ and $R_{2,3}=0.5$.

Case 2: $n=3$. Then $l=1$ and thus as above we get $R_{m, 3}=\frac{3 k}{3 m}>\frac{\sqrt{3} \cdot(m-\sqrt{3})}{3 m}=$ $\frac{\sqrt{3}}{3} \cdot\left(1-\frac{\sqrt{3}}{m}\right)$, which is greater than 0.5 for each $m>4 \sqrt{3}+6>12$, hence $m \in$ $\{3,4, \ldots, 12\}$. Direct calculations give $R_{3,3}=0.(3), R_{3,5}=0.4, R_{3,7}=4 / 7, R_{3,9}=$ $5 / 9, R_{3,11}=6 / 11$ and $R_{3,4}=R_{3,6}=R_{3,8}=R_{3,10}=R_{3,12}=0.5$.

Case 3: $n=4$. Then $l=2$ and thus as above we get $R_{m, 4}=\frac{6 k}{4 m}>\frac{\sqrt{3} \cdot(m-\sqrt{3})}{2 m}=$ $\frac{\sqrt{3}}{2} \cdot\left(1-\frac{\sqrt{3}}{m}\right)$, which is greater than 0.5 for each $m>\frac{\sqrt{27}+3}{2}>\frac{5+3}{2}=4$.

Hence $m=4$ and a calculation gives $R_{4,4}=0.75$.

Case 4: $n=5$. Then $l=2$ and thus as above we get $R_{m, 5}=\frac{6 k}{5 m}>\frac{2 \sqrt{3} \cdot(m-\sqrt{3})}{5 m}=$ $\frac{2 \sqrt{3}}{5} \cdot\left(1-\frac{\sqrt{3}}{m}\right)$, which is greater than 0.5 for each $m>\frac{12(4 \sqrt{3}+5)}{23}>\frac{12 \cdot 11}{23}>6$, hence $m \in\{5,6\}$. Direct calculations give $R_{5,5}=0.48$ and $R_{5,6}=0.6$.

We conclude that: $R_{i, j}=0.5$ for $(i, j) \in\{(2,3) ;(3,4) ;(3,6) ;(3,8) ;(3,10) ;(3,12)\}$.

These pairs are the dimensions of all rectangles with the desired property.