\[\left(1 + \frac{1}{1+2^1}\right)\left(1+\frac{1}{1+2^2}\right)\left(1 + \frac{1}{1+2^3}\right)\cdots\left(1 + \frac{1}{1+2^{10}}\right)= \frac{m}{n},\] where $m$ and $n$ are relatively prime positive integers. Find $m + n$.
1. Define the function \( f(n) = \prod_{k=1}^{n} \left(1 + \frac{1}{1 + 2^k}\right) \). We need to find \( f(10) \).
2. Calculate the first few values of \( f(n) \) to identify a pattern:
   \[
   f(1) = \left(1 + \frac{1}{1 + 2^1}\right) = \left(1 + \frac{1}{3}\right) = \frac{4}{3}
   \]
   \[
   f(2) = \left(1 + \frac{1}{1 + 2^1}\right)\left(1 + \frac{1}{1 + 2^2}\right) = \frac{4}{3} \cdot \left(1 + \frac{1}{5}\right) = \frac{4}{3} \cdot \frac{6}{5} = \frac{8}{5}
   \]
   \[
   f(3) = \left(1 + \frac{1}{1 + 2^1}\right)\left(1 + \frac{1}{1 + 2^2}\right)\left(1 + \frac{1}{1 + 2^3}\right) = \frac{8}{5} \cdot \left(1 + \frac{1}{9}\right) = \frac{8}{5} \cdot \frac{10}{9} = \frac{16}{9}
   \]
3. Based on these calculations, we suspect that \( f(n) = \frac{2^{n+1}}{2^n + 1} \). We will prove this by induction.

**Base Case:**
   \[
   f(1) = \frac{4}{3} = \frac{2^{1+1}}{2^1 + 1}
   \]
   The base case holds.

**Inductive Step:**
   Assume \( f(k) = \frac{2^{k+1}}{2^k + 1} \) for some \( k \). We need to show that \( f(k+1) = \frac{2^{k+2}}{2^{k+1} + 1} \).

   \[
   f(k+1) = f(k) \cdot \left(1 + \frac{1}{1 + 2^{k+1}}\right)
   \]
   Substitute the inductive hypothesis:
   \[
   f(k+1) = \frac{2^{k+1}}{2^k + 1} \cdot \left(1 + \frac{1}{1 + 2^{k+1}}\right) = \frac{2^{k+1}}{2^k + 1} \cdot \frac{2^{k+1} + 2}{2^{k+1} + 1}
   \]
   Simplify the expression:
   \[
   f(k+1) = \frac{2^{k+1}}{2^k + 1} \cdot \frac{2(2^k + 1)}{2^{k+1} + 1} = \frac{2^{k+1} \cdot 2(2^k + 1)}{(2^k + 1)(2^{k+1} + 1)} = \frac{2^{k+2}}{2^{k+1} + 1}
   \]
   The inductive step holds, so by induction, \( f(n) = \frac{2^{n+1}}{2^n + 1} \) for all \( n \).

4. Calculate \( f(10) \):
   \[
   f(10) = \frac{2^{10+1}}{2^{10} + 1} = \frac{2^{11}}{2^{10} + 1} = \frac{2048}{1025}
   \]

5. Since \( m = 2048 \) and \( n = 1025 \) are relatively prime, we have \( m + n = 2048 + 1025 = 3073 \).

The final answer is \(\boxed{3073}\).