For positive integer $n$ we define $f(n)$ as sum of all of its positive integer divisors (including $1$ and $n$). Find all positive integers $c$ such that there exists strictly increasing infinite sequence of positive integers $n_1, n_2,n_3,...$ such that for all $i \in \mathbb{N}$ holds $f(n_i)-n_i=c$
1. **Define the function \( f(n) \):**
   The function \( f(n) \) is defined as the sum of all positive divisors of \( n \). For example, if \( n = 6 \), then the divisors of 6 are 1, 2, 3, and 6, so \( f(6) = 1 + 2 + 3 + 6 = 12 \).

2. **Consider the case \( c = 1 \):**
   If \( c = 1 \), we need to find an infinite sequence of positive integers \( n_1, n_2, n_3, \ldots \) such that \( f(n_i) - n_i = 1 \) for all \( i \in \mathbb{N} \). This simplifies to \( f(n_i) = n_i + 1 \).

   - For prime numbers \( p \), the divisors are 1 and \( p \). Therefore, \( f(p) = 1 + p = p + 1 \).
   - Hence, for any prime number \( p \), \( f(p) - p = 1 \).

   Since there are infinitely many prime numbers, we can take the sequence of all prime numbers \( p_1, p_2, p_3, \ldots \). This sequence is strictly increasing and satisfies \( f(p_i) - p_i = 1 \) for all \( i \in \mathbb{N} \).

3. **Consider the case \( c > 1 \):**
   If \( c > 1 \), we need to find positive integers \( n \) such that \( f(n) = n + c \).

   - Suppose \( n \) is not a prime. Then \( n \) has at least one divisor \( d \) such that \( 1 < d < n \).
   - The sum of the divisors of \( n \) is at least \( 1 + d + n \) (since \( d \) and \( n \) are divisors, and 1 is always a divisor).
   - Therefore, \( f(n) \geq 1 + d + n \).

   Since \( d \geq 2 \) (as \( d \) is a proper divisor), we have:
   \[
   f(n) \geq 1 + 2 + n = n + 3
   \]
   This implies that \( f(n) - n \geq 3 \).

   - For \( f(n) - n = c \) to hold, we need \( c \geq 3 \).
   - If \( c > 1 \), then \( n \) must be composite, and we need to check if there can be infinitely many such \( n \).

   Consider the inequality:
   \[
   f(n) > \sqrt{n} + n
   \]
   This implies:
   \[
   \sqrt{n} < c
   \]
   Therefore, \( n < c^2 \). Since \( c \) is fixed and greater than 1, there can be at most finitely many such \( n \) satisfying \( n < c^2 \).

   Hence, there cannot be an infinite sequence of such \( n \) for \( c > 1 \).

Conclusion:
The only value of \( c \) for which there exists a strictly increasing infinite sequence of positive integers \( n_1, n_2, n_3, \ldots \) such that \( f(n_i) - n_i = c \) for all \( i \in \mathbb{N} \) is \( c = 1 \).

The final answer is \( \boxed{c = 1} \)