let $m,n$ be natural number with $m>n$ . find all such pairs of $(m,n) $
such that

$gcd(n+1,m+1)=gcd(n+2,m+2) =..........=gcd(m, 2m-n) = 1 $
1. Let \( n = x \) and \( m = x + k \), where \( x, k \in \mathbb{N} \). This means \( m \) and \( n \) are separated by \( k \) units.
2. We need to find pairs \((m, n)\) such that:
   \[
   \gcd(n+1, m+1) = \gcd(n+2, m+2) = \ldots = \gcd(m, 2m - n) = 1
   \]
3. Substitute \( n = x \) and \( m = x + k \) into the gcd conditions:
   \[
   \gcd(x+1, x+k+1) = \gcd(x+2, x+k+2) = \ldots = \gcd(x+k, 2x+k) = 1
   \]
4. Notice that for each \( i \) from \( 1 \) to \( k \), we have:
   \[
   \gcd(x+i, x+k+i) = \gcd(x+i, k)
   \]
   This follows from the property of gcd that \(\gcd(a, b) = \gcd(a, b - a)\).
5. For the gcd to be 1 for all \( i \) from \( 1 \) to \( k \), \( k \) must be coprime with each \( x+i \). However, since \( k \) is a fixed number, there will always be at least one \( i \) such that \( \gcd(x+i, k) \neq 1 \) unless \( k = 1 \).
6. If \( k = 1 \), then \( m = x + 1 \) and \( n = x \). This means \( m = n + 1 \).
7. Therefore, the only pairs \((m, n)\) that satisfy the given conditions are those where \( m = n + 1 \).

Conclusion:
\[
\boxed{(m, n) = (n+1, n) \text{ for all } n \in \mathbb{N}}
\]