Three consecutive integers are to be found, which have the following property: If we divide each pair of numbers in all possible ways, then the sum of the six resulting quotients is equal to an integer.

 untranslated portion: 

将上面的文本翻译成英文，请保留源文本的换行和格式，直接输出翻译结果。 

The above text has been translated into English, please keep the original text's line breaks and format, and output the translation result directly.
Let the three numbers be $x, x+1, x+2$.

Then

$$
\begin{gathered}
\frac{x}{x+1}+\frac{x}{x+2}+\frac{x+1}{x}+\frac{x+1}{x+2}+\frac{x+2}{x}+\frac{x+2}{x+1}=E \\
2+\frac{2 x+3}{x}+\frac{2 x+1}{x+2}=E \\
2+\frac{4 x^{2}+6 x+6}{x^{2}+2 x}=E \\
6+\frac{6}{x^{2}+2 x}=E
\end{gathered}
$$

The left side of the equation is an integer only if $x^{2}+2 x$ is a divisor of 6. Let

$$
x^{2}+2 x=d
$$

where $d=1,2,3,6$

$$
x=-1 \pm \sqrt{1+d}
$$

$1+d$ is a perfect square if $d=3$

then

$$
\begin{gathered}
x=-1 \pm 2 \\
x^{\prime}=1 \\
x^{\prime \prime}=-3
\end{gathered}
$$

The three numbers we are looking for are either

$$
1,2,3
$$

or

$$
-3,-2,-1
$$

(Mihály Seidner, 8th grade student, Losoncz.)

The problem was also solved by. Ignácz Bergstein, Nyíregyháza; VII. class of the State Real School in Debrecen; Gusztáv Szabó, Győr; Miklós Malesevits, high school teacher, Zombor.