1. In the interior angles of an $n(n \geqslant 3)$-sided polygon, the maximum number of acute angles is $\qquad$ .
II, 1.3.
Let a polygon with $n$ sides have $k$ acute angles, then
$$
(n-2) \times 180^{\circ}<k \times
$$
$90^{\circ}+(n-k) \times 180^{\circ}$.
Thus, $k<4$.
As shown in the figure, construct a $60^{\circ}$ sector $A_{1} A_{2} A_{1}$. When $n=3$, $\triangle A_{1} A_{2} A_{3}$ has three interior angles, all of which are part of $A_{1} A_{2} \cdots A_{n-1} A_{n}$, where $\angle A_{1} A_{2} A_{3}$, $\angle A_{n-1} A_{n} A_{1}$, and $\angle A_{n} A_{1} A_{2}$ are all acute angles, hence the maximum value of $k$ is 3.