Find the largest possible value of $k$ for which $3^{11}$ is expressible as the sum of $k$ consecutive positive integers.
1. Let the \( k \) consecutive positive integers be \( a, a+1, a+2, \ldots, a+(k-1) \). The sum of these integers can be expressed as:
   \[
   a + (a+1) + (a+2) + \cdots + (a+k-1)
   \]
   This is an arithmetic series with the first term \( a \) and the last term \( a+(k-1) \).

2. The sum of an arithmetic series is given by:
   \[
   \text{Sum} = \frac{\text{number of terms}}{2} \times (\text{first term} + \text{last term})
   \]
   Therefore, the sum of our series is:
   \[
   \frac{k}{2} \times (a + (a+k-1)) = \frac{k}{2} \times (2a + k - 1)
   \]
   Simplifying, we get:
   \[
   \frac{k(2a + k - 1)}{2}
   \]

3. We are given that this sum equals \( 3^{11} \):
   \[
   \frac{k(2a + k - 1)}{2} = 3^{11}
   \]
   Multiplying both sides by 2 to clear the fraction:
   \[
   k(2a + k - 1) = 2 \cdot 3^{11}
   \]
   \[
   k(2a + k - 1) = 2 \cdot 3^{11}
   \]

4. We need to find the largest possible value of \( k \). Notice that \( k \) must be a divisor of \( 2 \cdot 3^{11} \). We will consider two cases: \( k \) is odd and \( k \) is even.

5. **Case 1: \( k \) is odd**

   If \( k \) is odd, then \( k \) must be a power of 3 because the only prime factor of \( 3^{11} \) is 3. The largest power of 3 less than or equal to \( 3^{11} \) is \( 3^5 = 243 \).

6. **Case 2: \( k \) is even**

   If \( k \) is even, let \( k = 2p \). Substituting \( k = 2p \) into the equation:
   \[
   2p(2a + 2p - 1) = 2 \cdot 3^{11}
   \]
   Dividing both sides by 2:
   \[
   p(2a + 2p - 1) = 3^{11}
   \]
   Here, \( p \) must be a power of 3. The largest power of 3 less than or equal to \( 3^{11} \) is \( 3^5 = 243 \). Therefore, \( k = 2p = 2 \cdot 243 = 486 \).

7. Comparing the maximum values of \( k \) from both cases, we find that the largest possible value of \( k \) is 486.

The final answer is \( \boxed{486} \).