Suppose $a_{1} < a_{2}< \cdots < a_{2024}$ is an arithmetic sequence of positive integers, and  $b_{1} <b_{2} < \cdots <b_{2024}$ is a geometric sequence of positive integers. Find the maximum possible number of integers that could appear in both sequences, over all possible choices of the two sequences.

[i]Ray Li[/i]
1. **Define the sequences**: Let \( a_n \) be an arithmetic sequence and \( b_n \) be a geometric sequence. Specifically, we have:
   \[
   a_n = a_1 + (n-1)d
   \]
   \[
   b_n = b_1 \cdot r^{n-1}
   \]
   where \( d \) is the common difference of the arithmetic sequence, and \( r \) is the common ratio of the geometric sequence.

2. **Set up the problem**: We need to find the maximum number of integers that can appear in both sequences. Suppose there are \( k \) common terms between the sequences. Let these common terms be \( c_1, c_2, \ldots, c_k \).

3. **Common terms in both sequences**: For each common term \( c_i \), there exist integers \( m_i \) and \( n_i \) such that:
   \[
   c_i = a_1 + (m_i - 1)d = b_1 \cdot r^{n_i - 1}
   \]
   Since \( a_1 \) and \( b_1 \) are positive integers, and \( d \) and \( r \) are positive integers, we need to find the maximum \( k \) such that the above equation holds for distinct \( m_i \) and \( n_i \).

4. **Example with specific sequences**: Consider the arithmetic sequence \( a_n = n \) and the geometric sequence \( b_n = 2^{n-1} \). The common terms are:
   \[
   1, 2, 4, 8, 16, 32, 64, 128, 256, 512, 1024
   \]
   These are the first 11 powers of 2, which are common to both sequences.

5. **General case analysis**: Suppose there are more than 11 common terms. Let \( c_1, c_2, \ldots, c_{12} \) be 12 common terms. Then:
   \[
   c_i = a_1 + (m_i - 1)d = b_1 \cdot r^{n_i - 1}
   \]
   for \( i = 1, 2, \ldots, 12 \).

6. **Prime factor analysis**: Let \( r = \frac{q}{p} \) where \( q \) and \( p \) are relatively prime positive integers with \( q > p \geq 1 \). Then:
   \[
   b_i = b_1 \left( \frac{q}{p} \right)^{i-1}
   \]
   For \( b_i \) to be an integer, \( b_1 \) must be a multiple of \( p^{i-1} \). Let \( b_1 = p^{2023} x \) for some integer \( x \). Then:
   \[
   b_i = x p^{2024-i} q^{i-1}
   \]

7. **Divisibility constraints**: For the common terms \( c_i \) to be in both sequences, the differences \( c_{i+1} - c_i \) must be divisible by \( d \). Specifically:
   \[
   d \mid c_{i+1} - c_i = x p^{2024-i_{i+1}} q^{i_{i+1}-1} - x p^{2024-i_i} q^{i_i-1}
   \]
   Simplifying, we get:
   \[
   d \mid x p^{2024-i_{i+1}} q^{i_i-1} (q^{i_{i+1}-i_i} - p^{i_{i+1}-i_i})
   \]

8. **Prime factor bounds**: For any prime \( r \) dividing \( q \):
   \[
   \nu_r(d) \leq \nu_r(x q^{i_i-1})
   \]
   Repeating this for all 12 terms, we find that:
   \[
   \nu_r(b_{i_{k+1}} - b_{i_k}) - \nu_r(d) \geq \nu_r(q^{k-1})
   \]
   Thus:
   \[
   b_{i_{k+1}} - b_{i_k} \geq d q^{k-1}
   \]

9. **Summing the differences**: Summing these differences for 12 terms:
   \[
   b_{i_{12}} - b_{i_1} \geq d \sum_{k=1}^{11} q^{k-1} = d (q^{11} - 1) / (q - 1)
   \]
   Since \( q \geq 2 \):
   \[
   b_{i_{12}} - b_{i_1} \geq d (2^{11} - 1) = 2047d
   \]
   But \( b_{i_{12}} - b_{i_1} \leq 2023d \), a contradiction.

10. **Conclusion**: Therefore, there cannot be more than 11 common terms. We have already shown that 11 common terms are possible.

The final answer is \(\boxed{11}\)