`(1)/(7)+(1)/(8)+(1)/(9)+(1)/(10)lt(1)/(8)-(1)/(9)+(1)/(10)+(1)/(n)`
For the above inequality. What is the greatest possible positive integer value of n?

Answer

To solve the inequality

\[ \frac{1}{7} + \frac{1}{8} + \frac{1}{9} + \frac{1}{10} < \frac{1}{8} - \frac{1}{9} + \frac{1}{10} + \frac{1}{n} \]

we will follow these steps:

### Step 1: Move all terms to one side
We start by moving all the terms on the right side to the left side of the inequality:

\[ \frac{1}{7} + \frac{1}{8} + \frac{1}{9} + \frac{1}{10} - \left(\frac{1}{8} - \frac{1}{9} + \frac{1}{10} + \frac{1}{n}\right) < 0 \]

This simplifies to:

\[ \frac{1}{7} + \frac{1}{9} + \frac{1}{n} < 0 \]

### Step 2: Combine like terms
Now, we simplify the left side:

The terms \(\frac{1}{8}\) and \(-\frac{1}{8}\) cancel each other out, and the terms \(\frac{1}{10}\) and \(-\frac{1}{10}\) also cancel out. Thus, we have:

\[ \frac{1}{7} + \frac{1}{9} < \frac{1}{n} \]

### Step 3: Find a common denominator
To combine \(\frac{1}{7}\) and \(\frac{1}{9}\), we find the least common multiple (LCM) of 7 and 9, which is 63:

\[ \frac{1}{7} = \frac{9}{63}, \quad \frac{1}{9} = \frac{7}{63} \]

Adding these gives:

\[ \frac{9}{63} + \frac{7}{63} = \frac{16}{63} \]

### Step 4: Set up the inequality
Now we can rewrite the inequality:

\[ \frac{16}{63} < \frac{1}{n} \]

### Step 5: Cross-multiply
Cross-multiplying gives us:

\[ 16n < 63 \]

### Step 6: Solve for n
Now, we divide both sides by 16:

\[ n < \frac{63}{16} \]

Calculating \(\frac{63}{16}\) gives:

\[ \frac{63}{16} = 3.9375 \]

### Step 7: Find the greatest integer value
Since \(n\) must be a positive integer, the greatest possible value of \(n\) is 3.

### Final Answer
Thus, the greatest possible positive integer value of \(n\) is:

\[ \boxed{3} \]