Find all positive integer $n$ , such that for all 35-element-subsets of $M=(1,2,3,...,50)$ ,there exists at least two different elements $a,b$ , satisfing : $a-b=n$ or $a+b=n$.
1. Let \( A \) be any 35-element subset of \( M = \{1, 2, 3, \ldots, 50\} \).
2. Assume by contradiction that there exists a subset \( A \) such that no \( a, b \in A \) exist satisfying \( a - b = n \) or \( a + b = n \).
3. If \( a \in A \), then \( n - a \notin A \) and \( a - n \notin A \). Let \( m \) be the number of elements \( c \) such that \( c \in M \) and \( c \notin A \).
4. If \( m > 15 \), then \( A \) cannot exist because \( A \) needs to have 35 elements, but only \( 50 - m < 35 \) elements can be in \( A \).
5. Initially, let \( m = 0 \). If we take \( a \in A \) such that \( a > n \), then \( a - n \notin A \) and \( a - n \in M \), so \( m \) becomes \( m + 1 \).
6. If \( a < n \), then \( n - a \notin A \) and \( n - a \in M \) if and only if \( n - a \leq 50 \). Thus, \( m \) becomes \( m + 1 \) if and only if \( n - a \leq 50 \).
7. To minimize \( m \), we need at least 20 numbers in \( A \) satisfying \( n - a > 50 \) or \( n > 50 + a \). Minimizing \( n \), we get \( n > 70 \).
8. By our contradiction, we get that \( n \leq 70 \). Therefore, any \( n \in [1, 70] \) will work.

The final answer is \( \boxed{ n \in [1, 70] } \).