Solution:

The answer is $n-3$.

Lemma. If $ABCD$ is a convex quadrilateral with $\angle BAD + \angle CBA \geq \pi$ and $\angle BAD + \angle ADC \geq \pi$ then $A$ is a Bohemian vertex of $ABCD$.

Proof. Let $E$ be the reflection of $A$ in $ABCD$. It is clearly seen that $E$ belongs to the halfplanes containing $C$ determined by lines $AB$ and $AD$. Since $\angle BAD + \angle CBA \geq \pi$ and $\angle BAD + \angle EBA = \pi$, point $E$ belongs to the (closed) halfplane containing points $A, D$ determined by the line $BC$. Analogously, using the assumption $\angle BAD + \angle ADC$ we infer that $E$ belongs to the closed halfplane containing points $A, B$ determined by the line $CD$.

Therefore $E$ lies inside or on the boundary of $ABCD$. Thus $A$ is Bohemian.

Consider a convex $n$-gon $A_{1} A_{2} \ldots A_{n}$. Choose any four vertices $A_{i}, A_{j}, A_{k}, A_{l}$ with $i < j < k < l$ as in the picture below. Consider quadrilateral $A_{i} A_{j} A_{k} A_{l}$. It is clear that one of the points $A_{i}, A_{j}, A_{k}, A_{l}$ satisfies the assumption of the lemma, let's say this point is $A_{i}$. We claim that $A_{i}$ satisfies the assumption of the lemma in quadrilateral $A_{i-1} A_{i} A_{i+1} A_{k}$. Observe that the point $X := A_{k}A_{i+1} \cap A_{i}A_{i-1}$ lies in the triangle bounded by lines $A_{k}A_{j}, A_{j}A_{i}$ and $A_{i}A_{l}$. So
$$\angle A_{k}A_{i+1}A_{i} + \angle A_{i+1}A_{i}A_{i-1} = \pi + \angle A_{k}XA_{i} \geq \pi$$
(Note: it may happen that $X$ does not exist. It happens iff $j = i+1, l = i-1$ and $A_{k}A_{j} \parallel A_{l}A_{i}$. In that case $\angle A_{k}A_{i+1}A_{i} + \angle A_{i+1}A_{i}A_{i-1} = \pi$.)

Analogously $\angle A_{i+1}A_{i}A_{i-1} + \angle A_{i}A_{i-1}A_{k} \geq \pi$. Using the lemma we conclude that $A_{i}$ is a Bohemian vertex of quadrilateral $A_{i-1}A_{i}A_{i+1}A_{k}$. This implies that $A_{i}$ is a Bohemian vertex of $A_{1}A_{2}\ldots A_{n}$ since the quadrilateral $A_{i-1}A_{i}A_{i+1}A_{k}$ is a subset of the $n$-gon and the reflection point is the same.

Therefore, amongst any four vertices of a convex $n$-gon there exists a Bohemian vertex. So, every $n$-gon has at least $n-3$ Bohemian vertices.

An example of a convex $n$-gon with exactly $n-3$ Bohemian vertices is the following: take any kite $A_{1}A_{2}A_{3}A_{4}$ with $A_{4}A_{1} = A_{1}A_{2} < A_{2}A_{3} = A_{3}A_{4}$ and place points $A_{5}, \ldots, A_{n}$ very close to $A_{1}$. Then $A_{2}, A_{3}, A_{4}$ are not Bohemian vertices of $A_{1}A_{2}\ldots A_{n}$.

Solution:

We present a sketch of an alternative proof of the fact that every $n$-gon has at least $n-3$ Bohemian vertices.

Observation 1. Let us place the polygon into a coordinate system in such a way that $A_{1} = [0,0]$, $A_{2} = [a, 0]$, $a > 0$ and the second coordinates of all the remaining vertices are positive. If all the remaining vertices $A_{2}, \ldots, A_{n}$ have their first coordinates between $0$ and $a$ (see picture below), it is easy to see that the only vertices that could be non-Bohemian are $A_{1}, A_{2}$, and the point with the strictly largest second coordinate (if such a vertex exists). So, in this case, there exist at least $n-3$ Bohemian vertices.

Observation 2. An affine transformation does not change anything, so the statement is proved for all polygons that lie between two parallel lines that go through two adjacent vertices, i.e., whenever there are two adjacent vertices with sum of their angles at most $180^{\circ}$.

Consider now any polygon $P = A_{1}A_{2}, \ldots, A_{n}$.

Observation 3. If there are two (non-adjacent) vertices $A_{i}, A_{j}$ and two parallel lines $p_{i}, p_{j}$ with $A_{i} \in p_{i}, A_{j} \in p_{j}$ such that the whole polygon lies between $p_{i}$ and $p_{j}$, then the diagonal $A_{i}A_{j}$ splits $P$ into two polygons of the type considered in Observation 2. By Observations 1 and 2, these two polygons have at most 4 non-Bohemian points together, $A_{i}, A_{j}$, and two more.

Observation 4. For any vertex $A_{i}$ there exist a vertex $A_{j}$ and two parallel lines $p_{i}$ and $p_{j}$ with $A_{i} \in p_{i}, A_{j} \in p_{j}$ such that the whole polygon lies between them. In fact, take a line $p_{i}$ such that $p_{i} \cap P = \{A_{i}\}$, then $A_{j}$ is the vertex with maximal distance from $p_{i}$, if there are two such vertices, change the direction of $p_{i}$ slightly to obtain a unique $A_{j}$.

Let $n = 4$. Then there exist two adjacent vertices with sum of their angles not larger than $180^{\circ}$, so, by Observation 2, any quadrilateral has at most 3 non-Bohemian vertices.

Let $n \geq 5$. By Observations 3 and 4, there exist at most 4 non-Bohemian vertices. So, at least one vertex is Bohemian, denote it by $A_{i}$. Then, by Observations 3 and 4, all the non-Bohemian vertices are contained in the quadruple $A_{i}, A_{j}$ and some other two vertices. Since $A_{i}$ is Bohemian, there are at most three non-Bohemian vertices and the proof is complete.

Solution:

We prove by induction that every $n$-gon has at least $n-3$ Bohemian vertices.

Step 1, $n=4$. We show that every quadrilateral $ABCD$ has at least one Bohemian vertex. We consider a triangle $ABC$. Then $D$ has to be in one of the areas $P_{1}, P_{2}, P_{3}, P_{4}$ (see picture below), otherwise, $ABCD$ would not be a convex quadrilateral. If $D$ were in $P_{2}$, then $B$ would be Bohemian. If $D$ were in $P_{3}$, it would be Bohemian. If $D$ were in $P_{1}$, then $A$ would be Bohemian and similarly if $D$ were in $P_{4}$, then $C$ would be Bohemian (the last two cases are not immediate but easy to prove).

Induction step. Consider an $n$-gon $P = A_{1}A_{2}\ldots A_{n}$ with $n \geq 5$. Let $P'$ be an $(n-1)$-gon obtained from $P$ by omitting one vertex different from $A_{1}$. Let $A_{1}'$ be the reflection of $A_{1}$ in $P$ and $A_{1}''$ the reflection of $A_{1}$ in $P'$. We show the following statement
$$\text{if } A_{1} \text{ is non-Bohemian in } P, \text{ then it is non-Bohemian also in } P'.$$
This statement is obvious if the omitted vertex is not adjacent to $A_{1}$ (since in this case $A_{1}' = A_{1}''$ and $P' \subset P$). So, let the omitted vertex be $A_{n}$ (the other neighbour $A_{2}$ can be done in the same way) and let us assume for contradiction that $A_{1}' \notin P$ and $A_{1}'' \in P'$. Let us observe that vectors $A_{n}A_{n-1}$ and $A_{1}'A_{1}''$ are equal. Let us discuss the possible position of $A_{n-1}$. If $A_{n-1} \in Q_{3} \cup Q_{4}$ (as in the picture below) then $A_{1}''$ lies below the line $A_{n}A_{n-1}$ while the whole polygon $P$ lies above this line, contradiction with $A_{1}'' \in P' \subset P$. If $A_{n-1} \in Q_{2}$, then $A_{1}' \in \triangle A_{2}A_{n}A_{n-1} \subset P$, contradiction. If $A_{n-1} \in Q_{1}$, then the new reflection $A_{1}'' \in Q_{2}$ and $A_{1}' \in \triangle A_{2}A_{n}A_{1}''$ and $\triangle A_{2}A_{n}A_{1}'' \subset P'$ since $A_{1}'' \in P'$. Therefore $A_{1}' \in P' \subset P$, contradiction. Statement (S) is proved.

Since (by the induction hypothesis) there are at most 3 non-Bohemian vertices in $P'$, there are at most 4 non-Bohemian vertices in $P$ (the three and the omitted one). Since $n \geq 5$ there is at least one Bohemian vertex in $P$. Assume now that $P'$ is obtained from $P$ by omitting a Bohemian vertex. Since there are at most 3 non-Bohemian vertices in $P'$, there are at most 3 non-Bohemian vertices in $P$ and the proof is complete.