Let $\mathbb{N}$ be the set of positive integers. Let $n\in \mathbb{N}$ and let $d(n)$ be the number of divisors of $n$. Let $\varphi(n)$ be the Euler-totient function (the number of co-prime positive integers with $n$, smaller than $n$).
 
Find all non-negative integers $c$ such that there exists $n\in\mathbb{N}$ such that \[ d(n) + \varphi(n) = n+c , \] and for such $c$ find all values of $n$ satisfying the above relationship.
To solve the problem, we need to find all non-negative integers \( c \) such that there exists \( n \in \mathbb{N} \) satisfying the equation:
\[ d(n) + \varphi(n) = n + c. \]

We will analyze the equation for different values of \( n \).

1. **Case \( n = 1 \):**
   \[
   d(1) = 1 \quad \text{and} \quad \varphi(1) = 1.
   \]
   Therefore,
   \[
   d(1) + \varphi(1) = 1 + 1 = 2.
   \]
   This implies:
   \[
   2 = 1 + c \implies c = 1.
   \]

2. **Case \( n > 1 \):**
   We need to analyze the inequality:
   \[
   \varphi(n) + d(n) \leq n + 1.
   \]
   This inequality holds because \( \varphi(n) \) counts the number of integers less than \( n \) that are coprime with \( n \), and \( d(n) \) counts the number of divisors of \( n \). The sum of these two quantities cannot exceed \( n + 1 \).

   We now consider the equality cases:
   \[
   \varphi(n) + d(n) = n + 1.
   \]
   This equality holds if and only if every integer \( m < n \) that is not coprime with \( n \) is a divisor of \( n \).

   - **If \( n \) is a prime number:**
     \[
     \varphi(n) = n - 1 \quad \text{and} \quad d(n) = 2.
     \]
     Therefore,
     \[
     \varphi(n) + d(n) = (n - 1) + 2 = n + 1.
     \]
     This satisfies the equation.

   - **If \( n = 4 \):**
     \[
     \varphi(4) = 2 \quad \text{and} \quad d(4) = 3.
     \]
     Therefore,
     \[
     \varphi(4) + d(4) = 2 + 3 = 5 = 4 + 1.
     \]
     This satisfies the equation.

   - **If \( n \) is a composite number and \( n \neq 4 \):**
     Suppose \( n = ab \) where \( a, b \in \mathbb{N} \), \( a \geq 2 \), and \( b \geq 3 \). Consider \( m = a(b-1) \). We have \( (m, n) = a > 1 \) and \( n \) is not divisible by \( m \), which contradicts the condition for equality.

3. **Case \( \varphi(n) + d(n) = n \):**
   This equality holds if and only if there is exactly one exception to the rule that every \( m < n \) that is not coprime with \( n \) is a divisor of \( n \).

   - **If \( n = 6 \):**
     \[
     \varphi(6) = 2 \quad \text{and} \quad d(6) = 4.
     \]
     Therefore,
     \[
     \varphi(6) + d(6) = 2 + 4 = 6.
     \]
     This satisfies the equation.

   - **If \( n = 8 \):**
     \[
     \varphi(8) = 4 \quad \text{and} \quad d(8) = 4.
     \]
     Therefore,
     \[
     \varphi(8) + d(8) = 4 + 4 = 8.
     \]
     This satisfies the equation.

   - **If \( n = 9 \):**
     \[
     \varphi(9) = 6 \quad \text{and} \quad d(9) = 3.
     \]
     Therefore,
     \[
     \varphi(9) + d(9) = 6 + 3 = 9.
     \]
     This satisfies the equation.

   - **If \( n = 12 \):**
     \[
     \varphi(12) = 4 \quad \text{and} \quad d(12) = 6.
     \]
     Therefore,
     \[
     \varphi(12) + d(12) = 4 + 6 = 10.
     \]
     This does not satisfy the equation \( \varphi(n) + d(n) = n \).

Conclusion:
The values of \( c \) and corresponding \( n \) are:
- \( c = 1 \) for \( n = 1 \), all prime numbers, and \( n = 4 \).
- \( c = 0 \) for \( n = 6, 8, 9 \).

The final answer is \( \boxed{ c = 0 } \) and \( c = 1 \).