Fill in the squares with natural numbers so that:

$\triangleright$ the sum of all the filled-in numbers is 44,

$\triangleright$ the sum of the numbers in each four-square quadrant is the same,

$\triangleright$ the smallest filled-in number is odd,

$\triangleright$ the number in the center of the square is a single-digit number.

|  | 7 |  |
| :--- | :--- | :--- |
| 8 |  | 4 |
|  | 2 |  |

(S. Bednářová)
Let $x$ be the number in the middle square and $a$ be the number in the top-left square. Then the sum of the numbers in the top-left quarter square is $a+7+8+x=a+x+15$.

Next, we determine the number in the top-right square. To make the sum of the numbers in the top-right quarter square the same as the previous one (i.e., $a+x+15$), the number in the top-right square must be $a+4$ (the numbers 7 and $x$ are in both squares, so the sum of the remaining two numbers must be $8+a$).

Similarly, we determine that the number in the bottom-left square is $a+5$ (the numbers 8 and $x$ are in both quarter squares - the top-left and the bottom-left, the sum of the remaining two numbers is $a+7$).

Finally, we fill in the number in the bottom-right square: $a+9$.

The sum of all the filled numbers is 44 according to the problem, hence:

$$
\begin{aligned}
x+a+(a+4)+(a+5)+(a+9) & =44, \\
x+4 a+18 & =44, \\
x+4 a & =26, \\
x & =26-4 a .
\end{aligned}
$$

The possible pairs of numbers $x, a$ are listed in the following table:

$$
\begin{array}{l|c|c|c|c|c|c}
a & 1 & 2 & 3 & 4 & 5 & 6 \\
\hline x & 22 & 18 & 14 & 10 & 6 & 2
\end{array}
$$

The smallest of the filled numbers must be odd. We also know that the number in the middle (i.e., $x$) should be a single-digit number. Only one pair satisfies these conditions: $a=5$ and $x=6$.

After filling in all the numbers, we get:

| 5 | 7 | 9 |
| :---: | :---: | :---: |
| 8 | 6 | 4 |
| 10 | 2 | 14 |