Find all infinite sequences of positive integers $\{a_{n}\}_{n\geq 1}$ satisfying the following condition : there exists a positive constant $c$ such that $\gcd(a_{m}+n,a_{n}+m)>c(m+n)$ holds for all pairs of positive integers $(m,n)$.
1. **Stage 0: Define the function \( f \)**
   Define a function \( f: \mathbb{N} \rightarrow \mathbb{N} \) by \( f(n) = a_n \) for all \( n \in \mathbb{N} \).

2. **Stage 1: Establish a lower bound for \( f(m) \)**
   We need to show that there is a constant \( C \) such that \( f(m) > Cm \) for all \( m \in \mathbb{N} \) and \( \gcd(f(m) + n, f(n) + m) > C(m+n) \) for all \( m, n \in \mathbb{N} \).

   By letting \( n = 1 \) in the given equation, we have:
   \[
   \gcd(f(m) + 1, f(1) + m) > C(m+1)
   \]
   This implies that \( f(m) + 1 \) is divisible by some number greater than \( C(m+1) \). Therefore, \( f(m) > C(m+1) - 1 \). By taking a sufficiently small \( C < c \), we can ensure that \( f(m) > Cm \).

   Let \( P(m, n) \) denote the assertion:
   \[
   \gcd(f(m) + n, f(n) + m) > C(m+n)
   \]
   for all \( m, n \in \mathbb{N} \).

3. **Stage 2: Prime divisibility condition**
   We will show that for sufficiently large primes \( p \), \( p \mid f(i) - f(j) \) implies \( p \mid i - j \).

   Suppose otherwise, and select some \( n > 100(f(i)f(j)ij+1)^{100} \) such that \( p \mid n + f(i) \).

   By \( P(n, i) \) and \( P(n, j) \), we have:
   \[
   \gcd(f(i) + n, f(n) + i) > C(n+i)
   \]
   and
   \[
   \gcd(f(j) + n, f(n) + j) > C(n+j)
   \]
   If \( p \nmid f(n) + i \), then:
   \[
   \gcd(f(i) + n, f(n) + i) \le \frac{f(i) + n}{p}
   \]
   However, \( \frac{f(i) + n}{p} < C(n+i) \) because \( p \) was sufficiently large. Hence, we must have \( p \mid f(n) + i \). Analogously, \( p \mid f(n) + j \) is also true, and hence \( p \mid i - j \).

4. **Stage 2b: Adaptation for any prime \( p \)**
   We can show that for any prime \( p \), there exists some positive integer \( k_p \) such that \( p^t \mid f(i) - f(j) \) implies \( p^{t - k_p} \mid i - j \).

5. **Stage 3: Upper bound for \( f(n) \)**
   We will show that there is some constant \( D \) such that \( f(n) \le Dn \) for all \( n \in \mathbb{N} \).

   Combining our work in Stages 2 and 2b, we can see that there exists some positive integer \( T \) such that \( |f(i) - f(j)| \) divides \( T|i-j| \) for all \( i, j \in \mathbb{N} \).

   This means that:
   \[
   f(n) \le f(1) + |f(n) - f(1)| \le f(1) + T(n-1)
   \]
   which easily yields the desired result.

6. **Stage 4: Linear relationship for large \( n \)**
   We will show that \( f(n+1) = f(n) + 1 \) for all sufficiently large positive integers \( n \).

   Take some arbitrary \( n > \frac{1}{C}(1000T)^{1000} \). Observe that \( Cn \le f(n) \le Tn \). From our previous work, we have that \( |f(n) - f(n+1)| \le T \).

   By \( P(n, n+1) \), we have:
   \[
   \gcd(f(n) + n+1, f(n+1) + n) > C(2n+1)
   \]
   By the Triangle Inequality, we have:
   \[
   |(f(n) + n + 1) - (f(n+1) + n)| \le T + 1
   \]
   which implies that \( f(n) + n + 1 = f(n+1) + n \) because \( C(2n+1) > T+1 \).

7. **Stage 5: General form of \( f(n) \)**
   Because of our result from Stage 4, we know that there exists a constant \( \ell \in \mathbb{Z} \) such that \( f(n) = n + \ell \) for all sufficiently large \( n \).

   We'll now show that \( f(t) = t + \ell \) for all \( t \in \mathbb{N} \). Consider some arbitrary \( t \in \mathbb{N} \). Then take some positive integer \( n \) large enough so that \( f(n) = n + \ell \) and \( Cn > |f(t) - t - \ell| \).

   We then have from \( P(t, n) \) that:
   \[
   \gcd(f(t) + n, n + t + \ell) > C(m+n)
   \]
   which yields that \( f(t) = t + \ell \) from our assumption on \( n \).

From the result of Stage 5, it's easily seen that all solutions are \( f(x) \equiv x + c \) for \( c \in \mathbb{Z}_{\ge 0} \). We are done. \( \blacksquare \)

The final answer is \( \boxed{ f(x) \equiv x + c } \) for \( c \in \mathbb{Z}_{\ge 0} \).