A sequence $a_1,a_2,\ldots$ of positive integers satisfies the following properties.[list][*]$a_1 = 1$
[*]$a_{3n+1} = 2a_n + 1$
[*]$a_{n+1}\ge a_n$
[*]$a_{2001} = 200$[/list]Find the value of $a_{1000}$.

[i]Note[/i]. In the original statement of the problem, there was an extra condition:[list][*]every positive integer appears at least once in the sequence.[/list]However, with this extra condition, there is no solution, i.e., no such sequence exists. (Try to prove it.) The problem as written above does have a solution.
To solve the problem, we need to find the value of \(a_{1000}\) given the sequence properties. Let's break down the steps:

1. **Understanding the Sequence Definition:**
   - \(a_1 = 1\)
   - \(a_{3n+1} = 2a_n + 1\)
   - \(a_{n+1} \ge a_n\)
   - \(a_{2001} = 200\)

2. **Exploring the Sequence:**
   We start by exploring the sequence using the given properties. We know:
   - \(a_1 = 1\)
   - For \(n = 1\), \(a_4 = 2a_1 + 1 = 2 \cdot 1 + 1 = 3\)
   - For \(n = 4\), \(a_{13} = 2a_4 + 1 = 2 \cdot 3 + 1 = 7\)

3. **General Form:**
   We observe that \(a_{3n+1} = 2a_n + 1\). Let's generalize this:
   - For \(k\) iterations, \(a_{3^k n + \frac{3^k - 1}{2}} = (2^k - 1)a_n + \frac{2^k - 1}{2}\)

4. **Finding \(a_8\):**
   Given \(a_{2001} = 200\), we need to find \(a_8\):
   - For \(k = 5\) and \(n = 8\), \(3^5 \cdot 8 + \frac{3^5 - 1}{2} = 2065\)
   - \(a_{2065} = 31a_8\)
   - Since \(a_{2065} \ge 200\), we have \(31a_8 \ge 200 \Rightarrow a_8 \ge \frac{200}{31} \approx 6.45\)
   - Since \(a_8\) is an integer, \(a_8 \ge 7\)

5. **Verifying \(a_8 = 7\):**
   - We know \(a_{13} = 7\), so \(a_8 = 7\)

6. **Finding \(a_{1000}\):**
   - For \(k = 4\) and \(n = 8\), \(3^4 \cdot 8 + \frac{3^4 - 1}{2} = 688\)
   - \(a_{688} = 15a_8 + 7 = 15 \cdot 7 + 7 = 105\)
   - Therefore, \(a_{1000} = 105\)

The final answer is \(\boxed{105}\)