Solution:
The answer is negative.
Suppose such a partitioning was possible (Figure 7). Then $A B + B C + C D + D A \in \mathbb{N}$. But $(A B + B C) + (C D + D A) > A C + A C \geq 2$, hence $A B + B C + C D + D A > 2$.

On the other hand, $A B + B C + C D + D A < (A N + N B) + (B P + P C) + (C Q + Q D) + (D M + M A) = 4$, hence $A B + B C + C D + D A = 3$.

Obviously one of the sums of the elements of $S_{1}$ and $S_{2}$ must be $1$ and the other $2$. Without any loss of generality, we may assume that the sum of the elements of $S_{1}$ is $1$ and the sum of the elements of $S_{2}$ is $2$. As $A B + B C > A C \geq 1$ we find that $S_{1} \neq \{A B, B C\}$. Similarly, $S_{1}$ cannot contain two adjacent sides of the quadrilateral $A B C D$. Therefore, without any loss of generality, we may assume that $S_{1} = \{A D, B C\}$ and $S_{2} = \{A B, C D\}$. Then $A D + B C = 1$ and $A B + C D = 2$.

We have $A D \cdot B C \leq \frac{1}{4} (A D + C B)^2 = \frac{1}{4}$ and $A B \cdot C D \leq \frac{1}{4} (A B + C D)^2 = 1$.

According to Ptolemy's inequality, we have
$$\frac{5}{4} = A C \cdot B D \leq A B \cdot C D + A D \cdot B C = \frac{1}{4} + 1 = \frac{5}{4}$$
hence we have equality all around, which means the quadrilateral $A B C D$ is cyclic, $A D = B C = \frac{1}{2}$ and $A B = C D = 1$, hence $A B C D$ is a rectangle of dimensions $1$ and $\frac{1}{2}$.

There are many different ways of proving that this configuration is not possible. For example:

- Suppose $A B C D$ is a rectangle with $A D = \frac{1}{2}$, $A B = 1$. Then we have $A C = B D = \frac{\sqrt{5}}{2}$ and $\triangle A N B \equiv \triangle C Q D$ (Angle-Side-Angle). Denoting $A M = x$, $M D = y$ we have $A N =$

Figure 7: Exercise G7.

$1 - x$, $B N = 1 - y$ and the following conditions need to be fulfilled for some $x, y \in [0, 1]$ (Pythagorean Theorem in triangles $A M D$, $A N B$, $B B' C$, where $B'$ is the projection of $B$ on $M Q$):
$$x^2 + y^2 = \frac{1}{4}, \quad (1 - x)^2 + (1 - y)^2 = 1 \text{ and } 1 + (2y - 1)^2 = \frac{5}{4}$$
But $1 + (2y - 1)^2 = \frac{5}{4}$ implies $y \in \left\{\frac{1}{4}, \frac{3}{4}\right\}$. If $y = \frac{3}{4}$, then $x^2 + y^2 = \frac{1}{4}$ cannot hold. If on the other hand $y = \frac{1}{4}$, then $(1 - x)^2 + (1 - y)^2 = 1$ implies $x = 0$, but then $(1 - x)^2 + (1 - y)^2 = 1$ cannot hold. Therefore such a configuration is not possible.