If $n$ is an integer such that $n \ge 2^k$ and $n < 2^{k+1}$, where $k = 1000$, compute the following:
$$n - \left(  \lfloor \frac{n -2^0}{2^1} \rfloor + \lfloor \frac{n -2^1}{2^2} \rfloor + ...+ \lfloor \frac{n -2^{k-1}}{2^k} \rfloor \right)$$
To solve the given problem, we need to compute the expression:
\[ n - \left( \left\lfloor \frac{n - 2^0}{2^1} \right\rfloor + \left\lfloor \frac{n - 2^1}{2^2} \right\rfloor + \cdots + \left\lfloor \frac{n - 2^{k-1}}{2^k} \right\rfloor \right) \]
where \( n \) is an integer such that \( 2^k \le n < 2^{k+1} \) and \( k = 1000 \).

1. **Understanding the Floor Function Terms:**
   Each term \(\left\lfloor \frac{n - 2^i}{2^{i+1}} \right\rfloor\) counts the number of multiples of \(2^{i+1}\) that are less than or equal to \(n - 2^i\). This can be interpreted as the number of times \(2^{i+1}\) fits into \(n - 2^i\).

2. **Simplifying the Expression:**
   Let's rewrite the sum inside the floor functions:
   \[
   \sum_{i=0}^{k-1} \left\lfloor \frac{n - 2^i}{2^{i+1}} \right\rfloor
   \]
   We need to understand how many times each power of 2 fits into \(n\).

3. **Counting Multiples:**
   For each \(i\), the term \(\left\lfloor \frac{n - 2^i}{2^{i+1}} \right\rfloor\) essentially counts the number of multiples of \(2^{i+1}\) up to \(n - 2^i\). Since \(n\) is between \(2^k\) and \(2^{k+1}\), we can see that:
   \[
   \left\lfloor \frac{n - 2^i}{2^{i+1}} \right\rfloor = \left\lfloor \frac{n}{2^{i+1}} \right\rfloor - 1
   \]
   because \(2^i\) is subtracted from \(n\) and then divided by \(2^{i+1}\).

4. **Summing the Terms:**
   Summing these terms from \(i = 0\) to \(i = k-1\), we get:
   \[
   \sum_{i=0}^{k-1} \left( \left\lfloor \frac{n}{2^{i+1}} \right\rfloor - 1 \right)
   \]
   This simplifies to:
   \[
   \sum_{i=0}^{k-1} \left\lfloor \frac{n}{2^{i+1}} \right\rfloor - \sum_{i=0}^{k-1} 1
   \]
   The second sum is simply \(k\), since there are \(k\) terms.

5. **Final Expression:**
   Therefore, the expression inside the parentheses becomes:
   \[
   \sum_{i=0}^{k-1} \left\lfloor \frac{n}{2^{i+1}} \right\rfloor - k
   \]
   Substituting this back into the original expression, we get:
   \[
   n - \left( \sum_{i=0}^{k-1} \left\lfloor \frac{n}{2^{i+1}} \right\rfloor - k \right)
   \]
   Simplifying further:
   \[
   n - \sum_{i=0}^{k-1} \left\lfloor \frac{n}{2^{i+1}} \right\rfloor + k
   \]

6. **Conclusion:**
   The final expression simplifies to:
   \[
   n - \sum_{i=0}^{k-1} \left\lfloor \frac{n}{2^{i+1}} \right\rfloor + k
   \]

The final answer is \(\boxed{n - \sum_{i=0}^{k-1} \left\lfloor \frac{n}{2^{i+1}} \right\rfloor + k}\).