Find the positive integer $n$ such that $\frac12 \cdot \frac34 + \frac56 \cdot \frac78 +   \frac{9}{10}\cdot \frac{11}{12 }= \frac{n}{1200}$ .
1. First, we need to simplify each term in the given expression:
   \[
   \frac{1}{2} \cdot \frac{3}{4} = \frac{1 \cdot 3}{2 \cdot 4} = \frac{3}{8}
   \]
   \[
   \frac{5}{6} \cdot \frac{7}{8} = \frac{5 \cdot 7}{6 \cdot 8} = \frac{35}{48}
   \]
   \[
   \frac{9}{10} \cdot \frac{11}{12} = \frac{9 \cdot 11}{10 \cdot 12} = \frac{99}{120}
   \]

2. Next, we need to find a common denominator to add these fractions. The least common multiple (LCM) of 8, 48, and 120 is 240.

3. Convert each fraction to have a denominator of 240:
   \[
   \frac{3}{8} = \frac{3 \cdot 30}{8 \cdot 30} = \frac{90}{240}
   \]
   \[
   \frac{35}{48} = \frac{35 \cdot 5}{48 \cdot 5} = \frac{175}{240}
   \]
   \[
   \frac{99}{120} = \frac{99 \cdot 2}{120 \cdot 2} = \frac{198}{240}
   \]

4. Add the fractions:
   \[
   \frac{90}{240} + \frac{175}{240} + \frac{198}{240} = \frac{90 + 175 + 198}{240} = \frac{463}{240}
   \]

5. We are given that this sum equals \(\frac{n}{1200}\). Therefore, we set up the equation:
   \[
   \frac{463}{240} = \frac{n}{1200}
   \]

6. Solve for \(n\) by cross-multiplying:
   \[
   463 \cdot 1200 = 240 \cdot n
   \]
   \[
   555600 = 240n
   \]
   \[
   n = \frac{555600}{240} = 2315
   \]

The final answer is \(\boxed{2315}\).