4.1.11 * Find the integer pairs $(x, y)$ that satisfy the system of inequalities
$$
\left\{\begin{array}{l}
y-\left|x^{2}-2 x\right|+\frac{1}{2}>0, \\
y+|x-1|<2
\end{array}\right.
$$
From (1) we get $y>\left|x^{2}-2 x\right|-\frac{1}{2} \geqslant-\frac{1}{2}$. From (2) we get $y<2-|x-1| \leqslant 2$. Therefore, $-\frac{1}{2}<y<2$. The integers that satisfy this condition are $y=0, y=1$.
When $y=0$, we have $\left\{\begin{array}{l}\left|x^{2}-2 x\right|<\frac{1}{2}, \\ |x-1|<2 .\end{array}\right.$ From $|x-1|<2$, we get $-1<x<3$, so $x=0$, 1, 2. But $x-1$ does not satisfy $\left|x^{2}-2 x\right|<\frac{1}{2}$, so in this case, the integer solutions are $(x, y)=(0,0),(2,0)$.
When $y=1$, we have $\left\{\begin{array}{l}\left|x^{2}-2 x\right|<\frac{3}{2}, \\ |x-1|<1 .\end{array}\right.$ From $|x-1|<1$, we get $0<x<2$, so $x=1$. And $x=1$ also satisfies $\left|x^{2}-2 x\right|<\frac{3}{2}$, so in this case, the integer solution is $(x, y)=(1,1)$.
Therefore, the integer solutions $(x, y)$ that satisfy the original system of inequalities are $(0,0),(1,1),(2,0)$.