We claim that the answer is no. Proof: Let $a = x_1 + y_1$, $b = x_2 + y_2$, $c = x_3 + y_3$, and $d = x_4 + y_4$, such that $x_i \in \mathbb{Z}$, $y_i \in \mathbb{R}$, and $0 < y_i < 1$ for all $i = 1, 2, 3, 4$. Then we must have that the sum of each three of $y_1, y_2, y_3, y_4$ is an integer. If $y_1 + y_2 + y_3 \in \mathbb{Z}$ and $y_2 + y_3 + y_4 \in \mathbb{Z}$, then we must have $y_1 - y_4 \in \mathbb{Z}$. However, $-1 < y_1 - y_4 < 1$ so $y_1 - y_4 = 0$. Therefore $y_1 = y_4$ and we can continue similarly to obtain that $y_1 = y_2 = y_3 = y_4$. Then $3y_1 \in \mathbb{Z}$ and since $0 < y_1 < 1$ then $0 < 3y_1 < 3$ so $3y_1 = 1, 2$, and $y_1 = \frac{1}{3}, \frac{2}{3}$. Now suppose that $ab + cd \in \mathbb{Z}$. Then $ab + cd = (x_1 + y_1)(x_2 + y_2) + (x_3 + y_3)(x_4 + y_4) = (x_1 + y_1)(x_2 + y_1) + (x_3 + y_1)(x_4 + y_1) = x_1x_2 + x_3x_4 + y_1(x_1 + x_2 + x_3 + x_4 + 2y_1) \in \mathbb{Z}$. Then $y_1(x_1 + x_2 + x_3 + x_4 + 2y_1) \in \mathbb{Z}$. If $y_1 = \frac{1}{3}$ then $y_1(x_1 + x_2 + x_3 + x_4 + 2y_1) = \frac{x_1 + x_2 + x_3 + x_4 + \frac{2}{3}}{3} = \frac{3(x_1 + x_2 + x_3 + x_4) + 2}{9} \in \mathbb{Z}$. But since $3(x_1 + x_2 + x_3 + x_4) + 2 \equiv 2, 5, 8 \bmod 9$, this is impossible. If $y_1 = \frac{2}{3}$ then $y_1(x_1 + x_2 + x_3 + x_4 + 2y_1) = \frac{2(x_1 + x_2 + x_3 + x_4) + \frac{8}{3}}{3} = \frac{6(x_1 + x_2 + x_3 + x_4) + 8}{9} \in \mathbb{Z}$. But since $6(x_1 + x_2 + x_3 + x_4) + 8 \equiv 2, 5, 8 \bmod 9$, this is also impossible. Therefore, a contradiction has been reached, and we have proven that $ab + cd \notin \mathbb{Z}$.