Bára wrote down two different integers. Then she added, subtracted, multiplied, and divided them (in the same order). She got four results, whose sum was -100. When she omitted the result of the addition and added the remaining three results, she also got a sum of -100. What numbers could Bára have originally written?
Let the sought integers be $a, b$. According to the problem, two equations hold:

$$
\begin{gathered}
(a+b)+(a-b)+a b+(a: b)=-100 \\
(a-b)+a b+(a: b)=-100
\end{gathered}
$$

If the sum remains unchanged after omitting the result of addition, it is clear that $a+b=0$, i.e., $a=-b$. Substituting into the second equation, we get:

$$
-b-b-b^{2}-1=-100, \quad \text { i.e., }-(b+1)^{2}=-100 .
$$

Thus, $b+1= \pm 10$, so either $b=9$ (then $a=-9$), or $b=-11$ (then $a=11$). Therefore, Barbora could have written either the numbers -9 and 9, or the numbers 11 and -11.

$[a=-b \ldots 1$ point,

$-(b+1)^{2}=-100 \ldots 1$ point

calculation and answer $-9,9 \ldots 2$ points,

calculation and answer $11,-11 \ldots 2$ points]