We have calculated all possible two-number sums of four given numbers. Among them, the four smallest are 1, 5, 8, and 9. What were the four numbers?
Let the four numbers be arranged in increasing order as $a, b, c$, and $d$. Then, the size relationship of the six two-term sums is illustrated by the diagram. It is clear that, based solely on the condition $a \leq b \leq c \leq d$, the size relationship of any two sums can be determined except for the incomparable pair $a+d$ and $b+c$.

![](https://cdn.mathpix.com/cropped/2024_05_02_4e94e09d3ab76e00801bg-1.jpg?height=447&width=458&top_left_y=308&top_left_x=823)

The four smallest sums are therefore $a+b, a+c, a+d$, and $b+c$, and among them, the two smallest are, according to the condition,

$$
a+b=1 \quad \text{and} \quad a+c=5
$$

For the other two sums, we have either

(i) $a+d=8$ and $b+c=9$

or

(ii) $a+d=9$ and $b+c=8$.

Solving the systems of equations, in the first case we get

$$
a=-1.5 ; \quad b=2.5 ; \quad c=6.5 ; \quad d=9.5
$$

and in the second case, we get

$$
a=-1 ; \quad b=2 ; \quad c=6 ; \quad d=10
$$

Thus, there are two sets of four numbers that satisfy the conditions of the problem; these are listed in (1) and (2). Based on the work of Béla Kallós (Nyíregyháza, Krúdy Gy. Gymnasium, II. grade).