How many positive integers are there such that $\left \lfloor \frac m{11} \right \rfloor = \left \lfloor \frac m{10}  \right \rfloor$? ($\left \lfloor x \right \rfloor$ denotes the greatest integer not exceeding $x$.)

$ 
\textbf{(A)}\ 44
\qquad\textbf{(B)}\ 48
\qquad\textbf{(C)}\ 52
\qquad\textbf{(D)}\ 54
\qquad\textbf{(E)}\ 56
$
To solve the problem, we need to find the number of positive integers \( m \) such that \(\left \lfloor \frac{m}{11} \right \rfloor = \left \lfloor \frac{m}{10} \right \rfloor\).

1. **Define the floor functions:**
   Let \( k = \left \lfloor \frac{m}{11} \right \rfloor = \left \lfloor \frac{m}{10} \right \rfloor \). This implies:
   \[
   k \leq \frac{m}{11} < k+1 \quad \text{and} \quad k \leq \frac{m}{10} < k+1
   \]

2. **Translate the inequalities:**
   From the inequalities, we get:
   \[
   11k \leq m < 11(k+1) \quad \text{and} \quad 10k \leq m < 10(k+1)
   \]

3. **Combine the inequalities:**
   To satisfy both conditions simultaneously, \( m \) must lie in the intersection of the intervals:
   \[
   11k \leq m < 11(k+1) \quad \text{and} \quad 10k \leq m < 10(k+1)
   \]
   Therefore, \( m \) must satisfy:
   \[
   \max(11k, 10k) \leq m < \min(11(k+1), 10(k+1))
   \]
   Since \( 11k \geq 10k \), the combined inequality simplifies to:
   \[
   11k \leq m < 10(k+1)
   \]

4. **Determine the range of \( m \):**
   The range of \( m \) is from \( 11k \) to \( \min(11(k+1), 10(k+1)) - 1 \). Since \( 11(k+1) > 10(k+1) \), the upper bound is \( 10(k+1) - 1 \). Thus:
   \[
   11k \leq m < 10(k+1)
   \]

5. **Count the integers in the range:**
   The number of integers \( m \) in the range \( 11k \leq m < 10(k+1) \) is:
   \[
   10(k+1) - 11k = 10k + 10 - 11k = 10 - k
   \]

6. **Sum the counts for all \( k \):**
   We need to find the sum of the counts for all possible values of \( k \). The value of \( k \) ranges from 0 to the largest integer \( k \) such that \( 11k < 10(k+1) \), which simplifies to \( k < 10 \). Therefore, \( k \) ranges from 0 to 9.

   For each \( k \) from 0 to 9, the number of integers \( m \) is \( 10 - k \). Thus, we sum these counts:
   \[
   \sum_{k=0}^{9} (10 - k) = 10 + 9 + 8 + 7 + 6 + 5 + 4 + 3 + 2 + 1 = 55
   \]

The final answer is \(\boxed{55}\).