Juan chooses a five-digit positive integer.  Maria erases the ones digit and gets a four-digit number. The sum of this four-digit number and the original five-digit number is $52,713$. What can the sum of the five digits of the original number be?
1. Let the original five-digit number be represented as \(10x + y\), where \(x\) is a four-digit number and \(y\) is the units digit (0 through 9).
2. When Maria erases the ones digit, the resulting four-digit number is \(x\).
3. According to the problem, the sum of the original five-digit number and the four-digit number is \(52,713\). Therefore, we have:
   \[
   10x + y + x = 52,713
   \]
4. Simplifying the equation, we get:
   \[
   11x + y = 52,713
   \]
5. Since \(x\) is a four-digit number, it ranges from 1000 to 9999. We need to find values of \(x\) and \(y\) that satisfy the equation \(11x + y = 52,713\) where \(y\) is a digit (0 through 9).

6. Rearrange the equation to solve for \(y\):
   \[
   y = 52,713 - 11x
   \]
   Since \(y\) must be a digit (0 through 9), \(52,713 - 11x\) must yield a value in this range.

7. We now test values of \(x\) to find a valid \(y\):
   \[
   \frac{52,713 - y}{11} = x
   \]
   \(x\) must be an integer and a four-digit number.

8. Let's test \(y = 0\) through \(y = 9\):
   - For \(y = 0\):
     \[
     11x = 52,713 \implies x = \frac{52,713}{11} \approx 4792.0909 \quad (\text{not an integer})
     \]
   - For \(y = 1\):
     \[
     11x = 52,712 \implies x = \frac{52,712}{11} = 4792 \quad (\text{valid})
     \]
   - For \(y = 2\):
     \[
     11x = 52,711 \implies x = \frac{52,711}{11} \approx 4791.9091 \quad (\text{not an integer})
     \]
   - Continue this process for \(y = 3\) through \(y = 9\), and we find that only \(y = 1\) yields a valid integer \(x\).

9. Therefore, the original number is:
   \[
   10x + y = 10 \times 4792 + 1 = 47921
   \]

10. Finally, we calculate the sum of the digits of the original number \(47921\):
    \[
    4 + 7 + 9 + 2 + 1 = 23
    \]

The final answer is \(\boxed{23}\).