Croatian Mathematical Competitions

Let $a$, $b$, $c$ and $d$ be the numbers Azra thought of. Without loss of generality, we can assume that the deleted sum is $c+d$. Then there are numbers $a+b$, $a+c$, $a+d$, $b+c$ and $b+d$ written on the blackboard, i.e.
$$\{-2, 1, 2, 3, 6\} = \{a+b, a+c, a+d, b+c, b+d\}.$$
Since
$$(a+c) + (b+d) = (a+d) + (b+c),$$
among the numbers on the blackboard we can choose two pairs of numbers with equal sums. We easily find the only such pairs $-2+6=1+3=4$, and therefore the sum of all numbers Azra thought of equals $4$.
Among the numbers on the blackboard, the number $2$ does not appear in the last equality (so $a+b = 2$) and the deleted number $c+d$ equals $4-2=2$.
Let us now change the notation. Let $a$, $b$, $c$, $d$ be the numbers Azra thought of, such that $a \le b \le c \le d$. We know that $\{-2, 1, 2, 2, 3, 6\} = \{a+b, a+c, a+d, b+c, b+d, c+d\}$.
Since $a \le b \le c \le d$, obviously $a+b$ is the smallest sum, and $c+d$ is the largest, so $a+b = -2$, $c+d = 6$. Now it is easy to see that the sum $a+c$ is smaller than all the sums except $a+b$, and analogously $b+d$ is larger than all the sums except $c+d$. So, $a+c = 1$ and $b+d = 3$. Finally, $a+d = b+c = 2$.
We see that $2a = (a+b) + (a+c) - (b+c) = -2+1-2 = -3$, i.e. $a = -\frac{3}{2}$, and further $b = -2 - a = -\frac{1}{2}$, $c = 1 - a = \frac{5}{2}$ and $d = 2 - a = \frac{7}{2}$.
Let us check that the other equalities are satisfied: $b+c = -\frac{1}{2} + \frac{5}{2} = 2$, $b+d = -\frac{1}{2} + \frac{7}{2} = 3$, $c+d = \frac{5}{2} + \frac{7}{2} = 6$.
Azra thought of numbers $-\frac{3}{2}$, $-\frac{1}{2}$, $\frac{5}{2}$ and $\frac{7}{2}$.