Find the least positive integer  $n$ satisfying the following statement: for eash pair of positive integers $a$ and $b$ such that $36$ divides  $a+b$ and $n$ divides $ab$ it follows that $36$ divides both  $a$ and $b$.
To find the least positive integer \( n \) such that for each pair of positive integers \( a \) and \( b \) where \( 36 \) divides \( a + b \) and \( n \) divides \( ab \), it follows that \( 36 \) divides both \( a \) and \( b \), we proceed as follows:

1. **Understand the conditions:**
   - \( 36 \mid (a + b) \) implies \( a + b \equiv 0 \pmod{36} \).
   - \( n \mid ab \) implies \( ab \equiv 0 \pmod{n} \).

2. **Prime factorization of 36:**
   \[
   36 = 2^2 \times 3^2
   \]
   Therefore, for \( 36 \mid a \) and \( 36 \mid b \), both \( a \) and \( b \) must be divisible by \( 2^2 \) and \( 3^2 \).

3. **Analyze the divisibility conditions:**
   - If \( 36 \mid (a + b) \), then \( a + b \equiv 0 \pmod{36} \).
   - If \( n \mid ab \), then \( ab \equiv 0 \pmod{n} \).

4. **Determine the necessary conditions for \( n \):**
   - For \( 36 \mid a \) and \( 36 \mid b \), both \( a \) and \( b \) must be divisible by \( 4 \) and \( 9 \).
   - Therefore, \( ab \) must be divisible by \( 4 \times 4 = 16 \) and \( 9 \times 9 = 81 \).

5. **Find the least \( n \) such that \( n \mid ab \) implies \( 36 \mid a \) and \( 36 \mid b \):**
   - \( n \) must be at least \( 16 \times 81 = 1296 \).

6. **Verify the solution:**
   - If \( n = 1296 \), then \( 1296 = 2^4 \times 3^4 \).
   - If \( 1296 \mid ab \), then \( ab \) is divisible by \( 2^4 \) and \( 3^4 \).
   - Since \( 36 = 2^2 \times 3^2 \), both \( a \) and \( b \) must be divisible by \( 2^2 \) and \( 3^2 \) to ensure \( 36 \mid a \) and \( 36 \mid b \).

Therefore, the least positive integer \( n \) that satisfies the given conditions is \( 1296 \).

The final answer is \( \boxed{1296} \).