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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 $\frac{7}{2}$.