How many positive integer solutions does the equation have $$\left\lfloor\frac{x}{10}\right\rfloor= \left\lfloor\frac{x}{11}\right\rfloor + 1?$$
($\lfloor x \rfloor$ denotes the integer part of $x$,  for example $\lfloor 2\rfloor = 2$, $\lfloor \pi\rfloor = 3$, $\lfloor \sqrt2 \rfloor =1$)
To find the number of positive integer solutions to the equation 
\[ \left\lfloor\frac{x}{10}\right\rfloor= \left\lfloor\frac{x}{11}\right\rfloor + 1, \]
we will analyze the given floor functions and their properties.

1. **Initial Inequality Analysis:**
   \[ \frac{x}{10} - 1 < \left\lfloor\frac{x}{10}\right\rfloor = \left\lfloor\frac{x}{11}\right\rfloor + 1 \leq \frac{x}{11} + 1. \]
   This implies:
   \[ \frac{x}{10} - 1 < \frac{x}{11} + 1. \]
   Rearranging the terms, we get:
   \[ \frac{x}{10} - \frac{x}{11} < 2. \]
   Finding a common denominator:
   \[ \frac{11x - 10x}{110} < 2 \implies \frac{x}{110} < 2 \implies x < 220. \]

2. **Parameterization:**
   We can write \( x \) as:
   \[ x = 110k + \alpha, \]
   where \( 0 \le k \le 1 \) and \( 0 \le \alpha < 110 \).

3. **Substitution and Simplification:**
   Substituting \( x = 110k + \alpha \) into the equation:
   \[ \left\lfloor\frac{110k + \alpha}{10}\right\rfloor = \left\lfloor\frac{110k + \alpha}{11}\right\rfloor + 1. \]
   Simplifying the floor functions:
   \[ 11k + \left\lfloor\frac{\alpha}{10}\right\rfloor = 10k + \left\lfloor\frac{\alpha}{11}\right\rfloor + 1. \]
   Rearranging terms:
   \[ k + \left\lfloor\frac{\alpha}{10}\right\rfloor = \left\lfloor\frac{\alpha}{11}\right\rfloor + 1. \]

4. **Case Analysis:**
   We will consider two cases based on the value of \( k \).

   **Case 1: \( k = 0 \)**
   \[ \left\lfloor\frac{\alpha}{10}\right\rfloor = \left\lfloor\frac{\alpha}{11}\right\rfloor + 1. \]
   Let \( \alpha = 11a + b \) where \( 0 \le a \le 9 \) and \( 0 \le b \le 10 \). Then:
   \[ \left\lfloor\frac{11a + b}{10}\right\rfloor = \left\lfloor\frac{11a + b}{11}\right\rfloor + 1. \]
   Simplifying:
   \[ a + \left\lfloor\frac{a + b}{10}\right\rfloor = a + \left\lfloor\frac{b}{11}\right\rfloor + 1. \]
   \[ \left\lfloor\frac{a + b}{10}\right\rfloor = \left\lfloor\frac{b}{11}\right\rfloor + 1. \]
   Since \( 0 \le b \le 10 \), \( \left\lfloor\frac{b}{11}\right\rfloor = 0 \):
   \[ \left\lfloor\frac{a + b}{10}\right\rfloor = 1. \]
   This implies:
   \[ 10 \le a + b \le 19. \]
   For each \( a \) from 0 to 9, there are 10 possible values of \( b \) such that \( 10 \le a + b \le 19 \). Therefore, there are 55 such pairs.

   **Case 2: \( k = 1 \)**
   \[ 1 + \left\lfloor\frac{\alpha}{10}\right\rfloor = \left\lfloor\frac{\alpha}{11}\right\rfloor + 1. \]
   Simplifying:
   \[ \left\lfloor\frac{\alpha}{10}\right\rfloor = \left\lfloor\frac{\alpha}{11}\right\rfloor. \]
   Let \( \alpha = 11a + b \) where \( 0 \le a \le 9 \) and \( 0 \le b \le 10 \). Then:
   \[ \left\lfloor\frac{11a + b}{10}\right\rfloor = \left\lfloor\frac{11a + b}{11}\right\rfloor. \]
   Simplifying:
   \[ a + \left\lfloor\frac{a + b}{10}\right\rfloor = a + \left\lfloor\frac{b}{11}\right\rfloor. \]
   \[ \left\lfloor\frac{a + b}{10}\right\rfloor = \left\lfloor\frac{b}{11}\right\rfloor. \]
   Since \( 0 \le b \le 10 \), \( \left\lfloor\frac{b}{11}\right\rfloor = 0 \):
   \[ \left\lfloor\frac{a + b}{10}\right\rfloor = 0. \]
   This implies:
   \[ 0 \le a + b \le 9. \]
   For each \( a \) from 0 to 9, there are 10 possible values of \( b \) such that \( 0 \le a + b \le 9 \). Therefore, there are 55 such pairs.

5. **Total Solutions:**
   Adding the solutions from both cases:
   \[ 55 + 55 = 110. \]

The final answer is \(\boxed{110}\).