Claim: There can only be a maximum of $3$ acute angles in a convex polygon. Proof:We will continue with a proof by contradiction. Let $a_1, a_2, a_3, a_4$ be four acute angles in a convex polygon. We will prove that this cannot exist. We have that $a_1 + a_2 + a_3 + a_4 < 360^{\circ}$ since these angles are all acute. Assume we have some other angles $b_5, b_6, \dots, b_n$. Since there are only four acute angles in this polygon, the other angles must be right angles or obtuse angles (since this is a convex polygon, each of these angles is less than $180^{\circ}$). Thus, we have the following inequality: $$90^{\circ}(n-4) \leq b_5 + b_6 + \dots + b_n < 180^{\circ}(n-4).$$If we add this to the previous inequality, we get $$a_1 + \dots + a_4 + b_5 + \dots + b_n < 360^{\circ} + 180^{\circ}(n-4)$$Simplifying, we get $$a_1 + \dots + a_4 + b_5 + \dots + b_n < 180^{\circ}(n-2)$$However, since the total sum of angles in a convex polygon is $180^{\circ}(n-2)$, we have a contradiction. Thus, we cannot have $4$ acute angles in a convex polygon. $\square$ From this, we see that we can have a maximum of three acute angles in a convex polygon. It is trivial to construct convex polygons which meet this criteria (acute triangles have three acute angles, parallelograms which aren't rectangles have two acute angles, a "house-like" pentagon has one acute angle, and a regular hexagon has no acute angles).

Claim: There can only be a maximum of $3$ acute angles in a convex polygon. Proof:We will continue with a proof by contradiction. Let $a_1, a_2, a_3, a_4$ be four acute angles in a convex polygon. We will prove that this cannot exist. We have that $a_1 + a_2 + a_3 + a_4 < 360^{\circ}$ since these angles are all acute. Assume we have some other angles $b_5, b_6, \dots, b_n$. Since there are only four acute angles in this polygon, the other angles must be right angles or obtuse angles (since this is a convex polygon, each of these angles is less than $180^{\circ}$). Thus, we have the following inequality: $$90^{\circ}(n-4) \leq b_5 + b_6 + \dots + b_n < 180^{\circ}(n-4).$$If we add this to the previous inequality, we get $$a_1 + \dots + a_4 + b_5 + \dots + b_n < 360^{\circ} + 180^{\circ}(n-4)$$Simplifying, we get $$a_1 + \dots + a_4 + b_5 + \dots + b_n < 180^{\circ}(n-2)$$However, since the total sum of angles in a convex polygon is $180^{\circ}(n-2)$, we have a contradiction. Thus, we cannot have $4$ acute angles in a convex polygon. $\square$ From this, we see that we can have a maximum of three acute angles in a convex polygon. It is trivial to construct convex polygons which meet this criteria (acute triangles have three acute angles, parallelograms which aren't rectangles have two acute angles, a "house-like" pentagon has one acute angle, and a regular hexagon has no acute angles).

We look at the exterior lined angles. Let the number of sides of the polygon be $n$. Since we want to make the inner angle acute, we must make the other angle obtuse, and the sum of all the exterior lined angles is $360$, thus we must find the maximum natural $n$ such that $\frac{360}{n}>90$, which is $n=3$.