Twenty kilograms of cheese are on sale in a grocery store. Several customers are lined up to buy this cheese. After a while, having sold the demanded portion of cheese to the next customer, the salesgirl calculates the average weight of the portions of cheese already sold and declares the number of customers for whom there is exactly enough cheese if each customer will buy a portion of cheese of weight exactly equal to the average weight of the previous purchases. Could it happen that the salesgirl can declare, after each of the first $10$ customers has made their purchase, that there just enough cheese for the next $10$ customers? If so, how much cheese will be left in the store after the first $10$ customers have made their purchases? (The average weight of a series of purchases is the total weight of the cheese sold divided by the number of purchases.)
1. Let \( S_n \) be the sum of the weights of cheese bought after the \( n \)-th customer. We need to determine if the salesgirl can declare, after each of the first 10 customers, that there is just enough cheese for the next 10 customers if each customer buys a portion of cheese equal to the average weight of the previous purchases.

2. The total weight of cheese sold after \( n \) customers is \( S_n \). The average weight of the portions of cheese sold to the first \( n \) customers is \( \frac{S_n}{n} \).

3. The salesgirl declares that there is enough cheese for the next 10 customers if each of them buys a portion of cheese equal to the average weight of the previous purchases. This means that the remaining cheese, \( 20 - S_n \), should be equal to \( 10 \times \frac{S_n}{n} \).

4. Setting up the equation:
   \[
   20 - S_n = 10 \times \frac{S_n}{n}
   \]
   Simplifying, we get:
   \[
   20 - S_n = \frac{10S_n}{n}
   \]
   Multiplying both sides by \( n \) to clear the fraction:
   \[
   20n - nS_n = 10S_n
   \]
   Combining like terms:
   \[
   20n = 10S_n + nS_n
   \]
   \[
   20n = S_n (n + 10)
   \]
   Solving for \( S_n \):
   \[
   S_n = \frac{20n}{n + 10}
   \]

5. We need to check if the amount of cheese bought by any customer is positive. To do this, we calculate the difference between the cheese bought by the \( n \)-th customer and the \( (n-1) \)-th customer:
   \[
   S_n - S_{n-1} = \frac{20n}{n + 10} - \frac{20(n-1)}{(n-1) + 10}
   \]
   Simplifying the expression:
   \[
   S_n - S_{n-1} = 20 \left( \frac{n}{n + 10} - \frac{n-1}{n + 9} \right)
   \]
   Finding a common denominator:
   \[
   S_n - S_{n-1} = 20 \left( \frac{n(n + 9) - (n-1)(n + 10)}{(n + 10)(n + 9)} \right)
   \]
   Simplifying the numerator:
   \[
   n(n + 9) - (n-1)(n + 10) = n^2 + 9n - (n^2 + 10n - n - 10) = n^2 + 9n - n^2 - 9n + 10 = 10
   \]
   Thus:
   \[
   S_n - S_{n-1} = 20 \left( \frac{10}{(n + 10)(n + 9)} \right)
   \]
   Since the denominator is always positive for \( n \geq 1 \), \( S_n - S_{n-1} \) is positive, meaning the amount of cheese bought by any customer is positive.

6. To find the amount of cheese left in the store after the first 10 customers have made their purchases, we plug \( n = 10 \) into the expression for \( S_n \):
   \[
   S_{10} = \frac{20 \times 10}{10 + 10} = \frac{200}{20} = 10 \text{ kg}
   \]
   Therefore, the amount of cheese left in the store is:
   \[
   20 - S_{10} = 20 - 10 = 10 \text{ kg}
   \]

The final answer is \( \boxed{10} \) kg.