Suppose that \[\operatorname{lcm}(1024,2016)=\operatorname{lcm}(1024,2016,x_1,x_2,\ldots,x_n),\] with $x_1$, $x_2$, $\cdots$, $x_n$ are distinct postive integers. Find the maximum value of $n$.

[i]Proposed by Le Duc Minh[/i]
1. First, we need to find the least common multiple (LCM) of 1024 and 2016. We start by finding their prime factorizations:
   \[
   1024 = 2^{10}
   \]
   \[
   2016 = 2^5 \times 3^2 \times 7
   \]

2. The LCM of two numbers is found by taking the highest power of each prime that appears in their factorizations:
   \[
   \operatorname{lcm}(1024, 2016) = 2^{10} \times 3^2 \times 7
   \]

3. We are given that:
   \[
   \operatorname{lcm}(1024, 2016) = \operatorname{lcm}(1024, 2016, x_1, x_2, \ldots, x_n)
   \]
   This implies that the LCM of the set \(\{1024, 2016, x_1, x_2, \ldots, x_n\}\) must still be \(2^{10} \times 3^2 \times 7\).

4. Each \(x_i\) must be a factor of \(2^{10} \times 3^2 \times 7\) to ensure that the LCM does not change. We need to count the number of distinct factors of \(2^{10} \times 3^2 \times 7\).

5. The number of factors of a number \(p_1^{e_1} \times p_2^{e_2} \times \cdots \times p_k^{e_k}\) is given by:
   \[
   (e_1 + 1)(e_2 + 1) \cdots (e_k + 1)
   \]
   For \(2^{10} \times 3^2 \times 7\), the number of factors is:
   \[
   (10 + 1)(2 + 1)(1 + 1) = 11 \times 3 \times 2 = 66
   \]

6. Therefore, there are 66 distinct factors of \(2^{10} \times 3^2 \times 7\). Since \(1024\) and \(2016\) are already included in the set, the maximum number of additional distinct positive integers \(x_1, x_2, \ldots, x_n\) that can be included is \(66 - 2 = 64\).

Conclusion:
\[
\boxed{64}
\]