1. Understanding the Problem:
We need to prove that in a convex polygon partitioned into parallelograms, there are more than two vertices that belong to exactly one parallelogram (good vertices).

2. Lemma:
We start by considering the angles formed by the sides of the polygon and the parallelograms. Let the two marked angles be $\alpha$ and $\beta$. Since the parallelograms are non-intersecting, the sum of the angles at any vertex of the polygon must be at least $180^\circ$.

3. Angle Sum Analysis:
For a convex $n$-gon, the sum of the interior angles is given by:
$$180(n-2) \text{ degrees}$$
If we assume there are only two good vertices, then the remaining $n-2$ vertices must each belong to at least two parallelograms.

4. Contradiction Setup:
Assume for contradiction that there are only two good vertices. This means that the sum of the angles at the non-good vertices must be at least $180^\circ$ each. Summing these angles for all $n$ vertices, we get:
$$\sum_{i=1}^{n} \text{angle}_i \geq 180n$$
However, this sum includes the angles at the good vertices twice, so we need to subtract the excess.

5. Overcounting Adjustment:
The sum of the interior angles of the $n$-gon is:
$$180(n-2) \text{ degrees}$$
If we subtract the sum of the angles at the good vertices (which is less than $360^\circ$), we get:
$$180n - \text{sum of angles at good vertices} > 180(n-2)$$
This implies:
$$180n - \text{sum of angles at good vertices} > 180n - 360$$
Simplifying, we get:
$$\text{sum of angles at good vertices} < 360$$
This is a contradiction because the sum of the angles at the good vertices should be exactly $360^\circ$ if there are only two good vertices.

6. Conclusion:
Therefore, our assumption that there are only two good vertices must be false. Hence, there must be more than two good vertices in the convex polygon partitioned into parallelograms.

$$\blacksquare$$