1. Understanding the problem: We need to determine the smallest number of weights, $n$, such that Ana can measure any mass between $0$ and $20.2$ kg with an error of at most $10$ g. Each weight must be at least $10$ g.

2. Initial approach: Let's consider the weights $a_1, a_2, \ldots, a_n$. Each weight $a_i$ must be at least $10$ g. We need to ensure that any mass between $0$ and $20.2$ kg can be measured with an error of at most $10$ g.

3. **Checking $n = 2019$**:
- If we pick $a_1 = a_2 = \ldots = a_{2018} = 10 \frac{1}{10000}$ g and $a_{2019} = 20200 - 2018 \cdot 10 \frac{1}{10000} = 19 \frac{7972}{10000}$ g, the smallest possible sum of masses of weights is $10 \frac{1}{10000}$ g.
- This means the smallest measurable mass is $10 \frac{1}{10000}$ g, and we cannot measure masses smaller than this with the required precision. Hence, $n = 2019$ does not satisfy the requirements.

4. **Checking $n = 2020$**:
- If each weight is exactly $10$ g, then the total mass of the weights is $2020 \cdot 10 = 20200$ g or $20.2$ kg.
- This setup allows us to measure any mass in increments of $10$ g, from $0$ to $20.2$ kg.
- Since each weight is $10$ g, we can measure any mass with an error of at most $10$ g by determining which multiple of $10$ g the mass is closest to.

5. Conclusion: The smallest possible value of $n$ for which Ana can measure any mass between $0$ and $20.2$ kg with an error of at most $10$ g is $2020$.

The final answer is $\boxed{ n = 2020 }$.