1. Understanding the Problem:
We are given one million distinct points in the plane such that no three of them are collinear. We need to determine if it is possible to construct exactly ten nearly convex polygons using these points as vertices. A nearly convex polygon is defined as a non-self-intersecting polygon with exactly one interior angle greater than $180^\circ$.

2. Initial Assumptions:
Let's assume we can construct a nearly convex polygon $P$ with vertices $A_1, A_2, \ldots, A_{10^6}$ such that $\angle A_{10^6}A_1A_2 > 180^\circ$. This means that $A_1$ is a concave vertex.

3. Constructing Other Nearly Convex Polygons:
Consider the polygon $T = A_2A_3 \ldots A_{10^6}$. We need to show that $T$ is convex and that $A_1$ lies inside $T$. If $T$ is convex, then all its interior angles are less than $180^\circ$.

4. **Convexity of $T$:**
Since $T$ is formed by removing $A_1$ from $P$, and $P$ is nearly convex with only $\angle A_{10^6}A_1A_2 > 180^\circ$, the remaining angles in $T$ must be less than $180^\circ$. Therefore, $T$ is convex.

5. **Position of $A_1$:**
If $A_1$ lies outside $T$, then the line segment $A_1A_2$ would intersect $T$, which contradicts the assumption that $P$ is a non-self-intersecting polygon. Hence, $A_1$ must lie inside $T$.

6. Constructing Nearly Convex Polygons:
For each $i \in \{2, 3, \ldots, 10^6-1\}$, we can construct a nearly convex polygon $T_i$ by removing the segment $A_iA_{i+1}$ from $T$ and connecting $A_1$ to $A_i$ and $A_{i+1}$. This results in $10^6 - 1$ nearly convex polygons.

7. Conclusion:
Since we can construct $10^6 - 1$ nearly convex polygons, it is not possible to have exactly ten such polygons. The number of nearly convex polygons must be either zero or at least $10^6 - 1$.