Let $A$ be a positive real number. What are the possible values of $\displaystyle\sum_{j=0}^{\infty} x_j^2, $ given that $x_0, x_1, \cdots$ are positive numbers for which $\displaystyle\sum_{j=0}^{\infty} x_j = A$?
1. **Establishing the range of possible values:**
   - We are given that \( \sum_{j=0}^{\infty} x_j = A \) where \( x_j \) are positive numbers.
   - We need to determine the possible values of \( \sum_{j=0}^{\infty} x_j^2 \).

2. **Proving the upper bound:**
   - Since \( \sum_{j=0}^{\infty} x_j = A \), we can normalize the sequence by defining \( y_j = \frac{x_j}{A} \). Thus, \( \sum_{j=0}^{\infty} y_j = 1 \) and \( 0 < y_j < 1 \) for all \( j \).
   - Consider the sum of squares of the normalized sequence:
     \[
     \sum_{j=0}^{\infty} y_j^2
     \]
   - Since \( 0 < y_j < 1 \), it follows that \( y_j^2 < y_j \). Therefore,
     \[
     \sum_{j=0}^{\infty} y_j^2 < \sum_{j=0}^{\infty} y_j = 1
     \]
   - Multiplying through by \( A^2 \), we get:
     \[
     \sum_{j=0}^{\infty} x_j^2 = A^2 \sum_{j=0}^{\infty} y_j^2 < A^2
     \]
   - Hence, \( \sum_{j=0}^{\infty} x_j^2 \) is strictly less than \( A^2 \).

3. **Proving the lower bound:**
   - Since \( x_j \) are positive, \( x_j^2 \) are also positive. Therefore, \( \sum_{j=0}^{\infty} x_j^2 \) is positive.
   - Thus, \( 0 < \sum_{j=0}^{\infty} x_j^2 \).

4. **Constructing values within the interval \((0, A^2)\):**
   - To show that any value in the interval \((0, A^2)\) can be achieved, consider the sequence defined by \( x_j = A(1 - r)r^j \) for \( 0 < r < 1 \).
   - The sum of this geometric series is:
     \[
     \sum_{j=0}^{\infty} x_j = A \sum_{j=0}^{\infty} (1 - r)r^j = A \cdot \frac{1 - r}{1 - r} = A
     \]
   - Now, consider the sum of the squares:
     \[
     \sum_{j=0}^{\infty} x_j^2 = \sum_{j=0}^{\infty} \left(A(1 - r)r^j\right)^2 = A^2 (1 - r)^2 \sum_{j=0}^{\infty} r^{2j}
     \]
   - The sum of the geometric series \( \sum_{j=0}^{\infty} r^{2j} \) is:
     \[
     \sum_{j=0}^{\infty} r^{2j} = \frac{1}{1 - r^2}
     \]
   - Therefore,
     \[
     \sum_{j=0}^{\infty} x_j^2 = A^2 (1 - r)^2 \cdot \frac{1}{1 - r^2} = A^2 \cdot \frac{(1 - r)^2}{1 - r^2} = A^2 \cdot \frac{1 - r}{1 + r}
     \]
   - The expression \( \frac{1 - r}{1 + r} \) can take any value in \( (0, 1) \) by choosing an appropriate \( r \) in \( (0, 1) \). Specifically, if \( x = \frac{1 - r}{1 + r} \), then \( r = \frac{1 - x}{1 + x} \).

Conclusion:
- The set of possible values of \( \sum_{j=0}^{\infty} x_j^2 \) is the open interval \( (0, A^2) \).

The final answer is \( \boxed{ (0, A^2) } \).