Given is the equation $(m, n) +[m, n] =m+n$ where $m, n$ are positive integers and m>n. 
a) Prove that n divides m.
b) If $m-n=10$, solve the equation.
1. **Prove that \( n \) divides \( m \):**

   Given the equation \((m, n) + [m, n] = m + n\), where \( (m, n) \) denotes the greatest common divisor (gcd) of \( m \) and \( n \), and \([m, n]\) denotes the least common multiple (lcm) of \( m \) and \( n \).

   Let \( d = (m, n) \). Then we can write \( m = da \) and \( n = db \) where \( (a, b) = 1 \) (i.e., \( a \) and \( b \) are coprime).

   Substituting these into the given equation:
   \[
   d + \frac{m \cdot n}{(m, n)} = m + n
   \]
   \[
   d + \frac{da \cdot db}{d} = da + db
   \]
   \[
   d + dab = da + db
   \]
   \[
   d + dab = d(a + b)
   \]
   Dividing both sides by \( d \):
   \[
   1 + ab = a + b
   \]
   Rearranging the terms:
   \[
   ab - a - b + 1 = 0
   \]
   \[
   (a - 1)(b - 1) = 0
   \]
   Since \( a \) and \( b \) are positive integers and \( a > b \), it follows that \( b = 1 \).

   Therefore, \( m = da \) and \( n = d \). This implies that \( n \) divides \( m \) because \( m = da \) and \( n = d \).

   \(\blacksquare\)

2. **Solve the equation for \( m - n = 10 \):**

   Given \( m - n = 10 \), we substitute \( m = da \) and \( n = d \) into this equation:
   \[
   da - d = 10
   \]
   \[
   d(a - 1) = 10
   \]
   We need to find pairs \((d, a)\) such that \( d(a - 1) = 10 \). The factor pairs of 10 are:
   \[
   (d, a - 1) = (1, 10), (2, 5), (5, 2), (10, 1)
   \]
   Therefore, the corresponding pairs \((d, a)\) are:
   \[
   (d, a) = (1, 11), (2, 6), (5, 3), (10, 2)
   \]
   Substituting these back to find \((m, n)\):
   \[
   (m, n) = (1 \cdot 11, 1) = (11, 1)
   \]
   \[
   (m, n) = (2 \cdot 6, 2) = (12, 2)
   \]
   \[
   (m, n) = (5 \cdot 3, 5) = (15, 5)
   \]
   \[
   (m, n) = (10 \cdot 2, 10) = (20, 10)
   \]

The final answer is \( \boxed{ (m, n) = (11, 1), (12, 2), (15, 5), (20, 10) } \)