928. Solve in integers $x$ and $y$ the system of inequalities

$$
\left\{\begin{array}{l}
\left|x^{2}-2 x\right|<y+0.5 \\
y+|x-1|<2
\end{array}\right.
$$
$\triangle$ From the first inequality of the system, it is clear that $y \geqslant 0$, and from the second that $y<2$. Therefore, $0 \leqslant y<2$. Since only two integer values of $y$ satisfy the last inequality, $y=0$ and $y=1$, we will consider two cases.

1) Let $y=0$.

Then the system takes the form:

$$
\left\{\begin{array}{l}
\left|x^{2}-2 x\right|<0.5 \\
|x-1|<2
\end{array}\right.
$$

From this,

$$
\left\{\begin{array}{l}
-0.5<x^{2}-2 x<0.5 \\
-1<x<3
\end{array}\right.
$$

The last inequality is satisfied by the integers $x=0,1,2$. Substituting each of these values of $x$ into the first inequality, we establish that only $x=0$ and $x=2$ satisfy it. We obtain two solutions to the original system: $(0 ; 0)$ and $(2 ; 0)$.

2) Let $y=1$.

Then

$$
\left\{\begin{array}{l}
\left|x^{2}-2 x\right|<1.5, \\
|x-1|<1,
\end{array} \left\{\begin{array}{l}
-0.5<x^{2}-2 x<1.5 \\
0<x<2
\end{array}\right.\right.
$$

The last inequality is satisfied by only one integer $x: x=1$. It also satisfies the first inequality of the system. We obtain one more solution: $(1 ; 1)$.

Answer: $(0 ; 0),(2 ; 0),(1 ; 1)$.