Let $a_1,a_2,\ldots,a_n$ be the masses of the weights. For $n=2019$ pick $a_1 = a_2 = \ldots = a_{2018} = 10\frac{1}{10000}$ and $a_{2019} = 20200 - 2018 \cdot 10\frac{1}{10000} = 19\frac{7972}{10000}$. Then the smallest possible sum of masses of weights is $10\frac{1}{10000}$ and so the mass $\frac{1}{1000000}$ does not satisfy the requirements (even with one weight we would not know if the desired mass is less than $\frac{1}{10000}$ or greater than or equal to this quantity). Now we show that $n= 2020$ always works. We remind that each mass is at least $10$; so if any is strictly larger than $10$, then $20200 = \sum a_i > 2020 \cdot 10 = 20200$, contradiction. Hence all weights have mass exactly $10$, so all multiples of $10$ can be attained as total weight masses and now it is clear that any body weight can be determined between which consecutive multiples of $10$ it is.