Let the dimensions of the rectangle be $m \times n$, and WLOG $m \le n$. Claim: The maximum amount of squares of area $3$ (side length $\sqrt 3$) that can be put in the rectangle is $\lfloor \frac{m}{\sqrt{3}} \rfloor \lfloor \frac{n}{\sqrt{3}} \rfloor$. Proof: Let m = $a\sqrt3 + r_1$, and n = $b\sqrt3 + r_2$ $(a, b \in \mathbb{Z})$ Then we can fill the rectangle with $ab$ squares. Suppose FTSOC that there is space left for another square with area $3$. Then $a\sqrt3 + \sqrt3 \le m = a\sqrt3 + r_1$, absurd since $r_1 < \sqrt3$, so there is no space horizontally. The same goes for $b$ and $r_2$, so no space vertically either, which completes the proof. By the conditions of the problem, we obtain that $3\lfloor \frac{m}{\sqrt{3}} \rfloor \lfloor \frac{n}{\sqrt{3}} \rfloor = \frac{mn}{2}$. Then $2(m - \sqrt 3)(n - \sqrt 3) < mn$, or $mn + 6 < 2(m + n)\sqrt 3 < 4(m + n)$. Then $10 > (m - 4)(n - 4) \ge (m - 4)^2$ so $m \le 7$. Case 1: $m = 1$ $\implies 0 = \frac{n}{2}$ which is absurd. Case 2: $m = 2$ $\implies n = 3 \lfloor \frac{n}{\sqrt 3} \rfloor > (n - \sqrt 3)\sqrt 3$ or $3 > n(\sqrt 3 - 1) > \frac{n}{2}$ so $n < 6$ By testing $2 \le n < 6$ we obtain $n = 3$ as the only solution. Case 3: $m = 3$ $\implies n = 2 \lfloor \frac{n}{\sqrt 3} \rfloor > (n - \sqrt 3) \times \frac{2}{\sqrt 3}$ or $2 > n \times (2\sqrt3 - 1) > 2n$, absurd. Case 4: $m = 4$ $\implies n = 3\lfloor \frac{n}{\sqrt3} \rfloor$ so $4 \le n < 6$ which gives no solutions. Case 5: $m = 5$ $\implies 5n = 12\lfloor\frac{n}{\sqrt3}\rfloor > (n - \sqrt3) \times 4\sqrt3$ or $12 > n \times (4\sqrt3 - 5) > \frac{3n}{2}$ or $n > 8$ By testing $5 \le n < 8$ we obtain no solutions. Case 6: $m = 6$ $\implies 3\lfloor \frac{n}{\sqrt3} \rfloor = n$ so $6 \le n < 6$, absurd. Case 7: $m = 7$ $\implies 10 > 3(n - 4)$ or $n < \frac{22}{3}$ so $n = 7$ which isn't a solution. Therefore, the only rectangles that satisfy the property are $2 \times 3$ and $3 \times 2$ rectangles.