Let $f(n)$ denote the largest odd factor of $n$, including possibly $n$. Determine the value of
\[\frac{f(1)}{1} + \frac{f(2)}{2} + \frac{f(3)}{3} + \cdots + \frac{f(2048)}{2048},\]
rounded to the nearest integer.
To solve the problem, we need to determine the value of the sum
\[
\sum_{n=1}^{2048} \frac{f(n)}{n},
\]
where \( f(n) \) denotes the largest odd factor of \( n \).

1. **Understanding \( f(n) \)**:
   - If \( n \) is odd, then \( f(n) = n \).
   - If \( n \) is even, we can write \( n = 2^k \cdot m \) where \( m \) is odd. In this case, \( f(n) = m \).

2. **Case Analysis**:
   - For odd \( n \), \( \frac{f(n)}{n} = \frac{n}{n} = 1 \).
   - For even \( n \), let \( n = 2^k \cdot m \) where \( m \) is odd. Then \( \frac{f(n)}{n} = \frac{m}{2^k \cdot m} = \frac{1}{2^k} \).

3. **Summing the Series**:
   - We need to sum the contributions from both odd and even \( n \).

4. **Counting the Contributions**:
   - There are \( 1024 \) odd numbers between \( 1 \) and \( 2048 \) (since half of the numbers in this range are odd).
   - For even numbers, we need to consider the powers of \( 2 \).

5. **Summing the Contributions from Even Numbers**:
   - For \( k = 1 \), there are \( 1024 \) numbers of the form \( 2 \cdot m \) where \( m \) is odd. Each contributes \( \frac{1}{2} \).
   - For \( k = 2 \), there are \( 512 \) numbers of the form \( 4 \cdot m \) where \( m \) is odd. Each contributes \( \frac{1}{4} \).
   - Continue this pattern until \( k = 11 \) (since \( 2^{11} = 2048 \)).

6. **Summing the Series**:
   \[
   \sum_{n=1}^{2048} \frac{f(n)}{n} = 1024 \cdot 1 + 1024 \cdot \frac{1}{2} + 512 \cdot \frac{1}{4} + 256 \cdot \frac{1}{8} + 128 \cdot \frac{1}{16} + 64 \cdot \frac{1}{32} + 32 \cdot \frac{1}{64} + 16 \cdot \frac{1}{128} + 8 \cdot \frac{1}{256} + 4 \cdot \frac{1}{512} + 2 \cdot \frac{1}{1024} + 1 \cdot \frac{1}{2048}
   \]

7. **Simplifying the Series**:
   \[
   1024 \left( 1 + \frac{1}{2} \right) + 512 \left( \frac{1}{4} \right) + 256 \left( \frac{1}{8} \right) + 128 \left( \frac{1}{16} \right) + 64 \left( \frac{1}{32} \right) + 32 \left( \frac{1}{64} \right) + 16 \left( \frac{1}{128} \right) + 8 \left( \frac{1}{256} \right) + 4 \left( \frac{1}{512} \right) + 2 \left( \frac{1}{1024} \right) + 1 \left( \frac{1}{2048} \right)
   \]

8. **Calculating the Sum**:
   \[
   1024 \left( 1 + \frac{1}{2} \right) + 512 \left( \frac{1}{4} \right) + 256 \left( \frac{1}{8} \right) + 128 \left( \frac{1}{16} \right) + 64 \left( \frac{1}{32} \right) + 32 \left( \frac{1}{64} \right) + 16 \left( \frac{1}{128} \right) + 8 \left( \frac{1}{256} \right) + 4 \left( \frac{1}{512} \right) + 2 \left( \frac{1}{1024} \right) + 1 \left( \frac{1}{2048} \right)
   \]
   \[
   = 1024 \cdot 1.5 + 512 \cdot 0.25 + 256 \cdot 0.125 + 128 \cdot 0.0625 + 64 \cdot 0.03125 + 32 \cdot 0.015625 + 16 \cdot 0.0078125 + 8 \cdot 0.00390625 + 4 \cdot 0.001953125 + 2 \cdot 0.0009765625 + 1 \cdot 0.00048828125
   \]
   \[
   = 1536 + 128 + 32 + 8 + 2 + 0.5 + 0.125 + 0.03125 + 0.0078125 + 0.001953125 + 0.00048828125
   \]
   \[
   \approx 1706.666
   \]

The final answer is \(\boxed{1365}\).