Let $f: N \to  N$ satisfy $n=\sum_{d|n} f(d), \forall n \in N$. Then sum of all possible values of $f(100)$ is? 
1. **Restate the given condition using convolution:**
   The problem states that for a function \( f: \mathbb{N} \to \mathbb{N} \), the following holds:
   \[
   n = \sum_{d|n} f(d) \quad \forall n \in \mathbb{N}
   \]
   This can be restated using the concept of Dirichlet convolution. Let \(\text{id}(n) = n\) be the identity function. The given condition can be written as:
   \[
   \text{id} = f * 1
   \]
   where \( * \) denotes the Dirichlet convolution.

2. **Apply Möbius inversion:**
   By the Möbius inversion formula, if \( g = f * h \), then \( f = g * \mu \), where \(\mu\) is the Möbius function. In our case, \( g = \text{id} \) and \( h = 1 \), so:
   \[
   f = \text{id} * \mu
   \]

3. **Calculate \( f(100) \):**
   We need to find \( f(100) \). Using the formula \( f = \text{id} * \mu \), we have:
   \[
   f(n) = \sum_{d|n} \text{id}(d) \mu\left(\frac{n}{d}\right)
   \]
   For \( n = 100 \):
   \[
   f(100) = \sum_{d|100} d \mu\left(\frac{100}{d}\right)
   \]
   The divisors of 100 are \( 1, 2, 4, 5, 10, 20, 25, 50, 100 \). We calculate each term:
   \[
   \begin{aligned}
   &d = 1, & \mu(100) &= 0, & 1 \cdot 0 &= 0, \\
   &d = 2, & \mu(50) &= 0, & 2 \cdot 0 &= 0, \\
   &d = 4, & \mu(25) &= 0, & 4 \cdot 0 &= 0, \\
   &d = 5, & \mu(20) &= 0, & 5 \cdot 0 &= 0, \\
   &d = 10, & \mu(10) &= 1, & 10 \cdot 1 &= 10, \\
   &d = 20, & \mu(5) &= -1, & 20 \cdot (-1) &= -20, \\
   &d = 25, & \mu(4) &= 0, & 25 \cdot 0 &= 0, \\
   &d = 50, & \mu(2) &= -1, & 50 \cdot (-1) &= -50, \\
   &d = 100, & \mu(1) &= 1, & 100 \cdot 1 &= 100.
   \end{aligned}
   \]
   Summing these values:
   \[
   f(100) = 0 + 0 + 0 + 0 + 10 - 20 + 0 - 50 + 100 = 40
   \]

The final answer is \(\boxed{40}\)