7.26 For all the interior angles of a non-self-intersecting $n$-sided polygon in a plane, what is the maximum number of acute angles?
[Solution] Let there be $k$ acute angles in an $n$-sided polygon. Then we have
$$
k \times 90^{\circ} + (n-k) \times 360^{\circ} > (n-2) \times 180^{\circ}.
$$

Simplifying the inequality, we get
$$
3k < 2n + 4.
$$

Since both sides of the inequality are integers, we have
$$
3k \leqslant 2n + 3.
$$

This shows that the number of acute angles in an $n$-sided polygon does not exceed $\left[\frac{2n}{3}\right] + 1$.
Next, we use the construction method to show that $\left[\frac{2n}{3}\right] + 1$ can be achieved. First, consider the case where $n$ is a multiple of 3: $n = 3r$, so $\left[\frac{2n}{3}\right] + 1 = 2r + 1$. Take a sector with a central angle of $60^{\circ}$, with its vertex at $P$ and the endpoints of the arc at $A$ and $B$. Divide the arc $AB$ into $2r-1$ equal parts using points $C_1, C_2, \cdots, C_{2r-2}$. Let $S_i$ be the centroid of $\triangle C_{2i-1}PC_{2i}$, for $i = 1, 2, \cdots, r-1$. Draw lines through each point $S_i$ parallel to the radii $PC_{2i-1}$ and $PC_{2i}$, intersecting the extensions of the chords $C_{2i-2}C_{2i-1}$ and $C_{2i+1}C_{2i}$ at points $D_{2i-1}$ and $D_{2i}$, respectively. This way, we obtain an $n$-sided polygon $\sum = P A D_1 S_1 D_2 D_3 \cdots D_{2i-1} S_i D_{2i} \cdots D_{2r-2} B$ (see the figure). This $n$-sided polygon has $2r + 1$ acute angles, with their vertices at $P, A, D_1, D_2, \cdots, D_{2r-2}, B$.

When $n$ is 1 more than a multiple of 3: $n = 3r + 1$, $\left[\frac{2n}{3}\right] + 1 = 2r + 1$. In this case, we just need to replace the side $PA$ of the polygon $\sum$ with a broken line $PQA$ (as shown by the dashed line in the figure).

When $n = 3r + 2$, we construct a $\frac{1}{4}$ equilateral triangle $PQ_1Q_2$ with side length equal to the radius inside the sector, with vertex at $P$. Then, construct the symmetric figure $\triangle P_1Q_1Q_2$ of $\triangle PQ_1Q_2$ with $Q_1Q_2$ as the axis. Remove the segments $Q_2P$ and $Q_1P$ from the sides $BP$ and $AP$ of the polygon $\sum$ and replace them with the segments $Q_2P_1$ and $Q_1P_1$, respectively, to obtain a $(3r + 2)$-sided polygon with $2r + 2 = \left[\frac{2n}{3}\right] + 1$ acute angles.
In summary, an $n$-sided polygon has at most $\left[\frac{2n}{3}\right] + 1$ acute angles.