II. (25 points) On the coordinate plane: Points whose both coordinates are integers are called integer points. Try to find all integer points $(x, y)$ on the graph of the quadratic function $y=\frac{x^{2}}{10}-\frac{x}{10}+\frac{9}{5}$ that satisfy $y \leqslant|x|$, and explain your reasoning.
II. Solve $\because y \leqslant|x|$, i.e., $\frac{x^{2}-x+18}{10} \leqslant|x|$,
Thus, $x^{2}-x+18 \leqslant 10|x|$.
(*)
When $x \geqslant 0$, equation (*) becomes
$$
x^{2}-x+18 \leqslant 10 x \text {, }
$$

which simplifies to $x^{2}-11 x+18 \leqslant 0$.
Therefore, $2 \leqslant x \leqslant 9$.
In this case, the points that satisfy the condition are $(2,2),(4,3),(7,6),(9$,
$9)$.
When $x<0$, equation (*) becomes
$$
x^{2}-x+18 \leqslant-10 x \text {, }
$$

which simplifies to
$$
x^{2}+9 x+18 \leqslant 0 \text {. }
$$

Therefore, $-6 \leqslant x \leqslant-3$.
In this case, the points that satisfy the condition are $(-6,6),(-3,3)$. Thus, the points that satisfy the condition are the six integer points listed above.