For every natural number $a$ and $b$, define the notation $[a,b]$ as the least common multiple of $a $ and $b$ and the notation $(a,b)$ as the greatest common divisor of $a$ and $b$. Find all $n \in \mathbb{N}$ that satisfies
\[
4 \sum_{k=1}^{n} [n,k] = 1 + \sum_{k=1}^{n} (n,k) + 2n^2 \sum_{k=1}^{n} \frac{1}{(n,k)}
\]
To solve the given problem, we need to find all natural numbers \( n \) that satisfy the equation:
\[
4 \sum_{k=1}^{n} [n,k] = 1 + \sum_{k=1}^{n} (n,k) + 2n^2 \sum_{k=1}^{n} \frac{1}{(n,k)}
\]

1. **Expressing the Least Common Multiple and Greatest Common Divisor:**
   We know that for any two natural numbers \( a \) and \( b \),
   \[
   ab = (a, b) [a, b]
   \]
   Therefore, for \( n \) and \( k \),
   \[
   [n, k] = \frac{nk}{(n, k)}
   \]

2. **Substituting \( [n, k] \) in the Given Equation:**
   Substitute \( [n, k] = \frac{nk}{(n, k)} \) into the given equation:
   \[
   4 \sum_{k=1}^{n} \frac{nk}{(n, k)} = 1 + \sum_{k=1}^{n} (n,k) + 2n^2 \sum_{k=1}^{n} \frac{1}{(n,k)}
   \]

3. **Simplifying the Equation:**
   Simplify the left-hand side:
   \[
   4n \sum_{k=1}^{n} \frac{k}{(n, k)}
   \]
   The equation now becomes:
   \[
   4n \sum_{k=1}^{n} \frac{k}{(n, k)} = 1 + \sum_{k=1}^{n} (n,k) + 2n^2 \sum_{k=1}^{n} \frac{1}{(n,k)}
   \]

4. **Analyzing the Summation Terms:**
   Notice that \( (n, k) = (n, n-k) \). This symmetry can help simplify the summations. Consider the term:
   \[
   \sum_{k=1}^{n} \frac{2n(n - 2k)}{(n, k)}
   \]
   Observe that:
   \[
   \frac{2n(n - 2(n - k))}{(n, n - k)} = \frac{2n(2k - n)}{(n, k)} = -\frac{2n(n - 2k)}{(n, k)}
   \]
   Hence,
   \[
   \sum_{k=1}^{n} \frac{2n(n - 2k)}{(n, k)} = -2n
   \]

5. **Simplifying Further:**
   The equation reduces to:
   \[
   1 + \sum_{k=1}^{n} (n, k) - 2n = 0
   \]
   Rearrange to:
   \[
   \sum_{k=1}^{n} (n, k) = 2n - 1
   \]

6. **Case Analysis:**
   - **Case 1: \( n \) is prime:**
     If \( n \) is prime, then \( (n, k) = 1 \) for all \( k \) from 1 to \( n-1 \), and \( (n, n) = n \). Therefore,
     \[
     \sum_{k=1}^{n} (n, k) = 1 + 1 + \cdots + 1 + n = (n-1) + n = 2n - 1
     \]
     This satisfies the equation.

   - **Case 2: \( n \) is not prime:**
     If \( n \) is not prime, we need to show that:
     \[
     \sum_{k=1}^{n} (n, k) > 2n - 1
     \]
     For non-prime \( n \), there are divisors \( d \) such that \( 1 < d < n \). Hence, the sum \( \sum_{k=1}^{n} (n, k) \) will include terms greater than 1, making the sum larger than \( 2n - 1 \).

Therefore, the only solutions are when \( n \) is a prime number.

The final answer is \( \boxed{ n } \) is a prime number.