Let the sequence $(a_n)_{n \in \mathbb{N}}$, where $\mathbb{N}$ denote the set of natural numbers, is given with $a_1=2$ and $a_{n+1}$ $=$ $a_n^2$ $-$ $a_n+1$. Find the minimum real number $L$, such that for every $k$ $\in$ $\mathbb{N}$
\begin{align*} \sum_{i=1}^k \frac{1}{a_i} < L \end{align*}
1. **Claim 1:**
   We need to prove that \(\prod_{i=1}^j a_i = a_{j+1} - 1\).

   **Proof:**
   By the given recurrence relation, we have:
   \[
   a_{k+1} - 1 = a_k^2 - a_k = a_k(a_k - 1)
   \]
   We can continue this process:
   \[
   a_k(a_k - 1) = a_k(a_{k-1}^2 - a_{k-1}) = a_k a_{k-1}(a_{k-1} - 1)
   \]
   Repeating this process, we get:
   \[
   a_k a_{k-1} (a_{k-1} - 1) = a_k a_{k-1} a_{k-2} (a_{k-2} - 1) = \cdots = \prod_{i=1}^k a_i
   \]
   Therefore, we have:
   \[
   a_{k+1} - 1 = \prod_{i=1}^k a_i
   \]
   This completes the proof of Claim 1. \(\blacksquare\)

2. **Claim 2:**
   We need to show that the sum \(\sum_{i=1}^k \frac{1}{a_i}\) is of the form \(\frac{\prod_{i=1}^k a_i - 1}{\prod_{i=1}^k a_i}\).

   **Proof:**
   We will use induction on \(k\).

   - **Base Case:**
     For \(k=1\):
     \[
     \sum_{i=1}^1 \frac{1}{a_i} = \frac{1}{a_1} = \frac{1}{2}
     \]
     which is of the form \(\frac{a_1 - 1}{a_1} = \frac{2 - 1}{2} = \frac{1}{2}\).

     For \(k=2\):
     \[
     \sum_{i=1}^2 \frac{1}{a_i} = \frac{1}{a_1} + \frac{1}{a_2} = \frac{1}{2} + \frac{1}{5} = \frac{5 + 2}{10} = \frac{7}{10}
     \]
     which is of the form \(\frac{\prod_{i=1}^2 a_i - 1}{\prod_{i=1}^2 a_i} = \frac{10 - 1}{10} = \frac{9}{10}\).

   - **Induction Step:**
     Assume the statement is true for \(k = g\), i.e.,
     \[
     \sum_{i=1}^g \frac{1}{a_i} = \frac{\prod_{i=1}^g a_i - 1}{\prod_{i=1}^g a_i}
     \]
     We need to show it is true for \(k = g+1\):
     \[
     \sum_{i=1}^{g+1} \frac{1}{a_i} = \sum_{i=1}^g \frac{1}{a_i} + \frac{1}{a_{g+1}}
     \]
     Using the induction hypothesis:
     \[
     \sum_{i=1}^{g+1} \frac{1}{a_i} = \frac{\prod_{i=1}^g a_i - 1}{\prod_{i=1}^g a_i} + \frac{1}{a_{g+1}}
     \]
     Combining the fractions:
     \[
     \sum_{i=1}^{g+1} \frac{1}{a_i} = \frac{(\prod_{i=1}^g a_i - 1) a_{g+1} + 1}{\prod_{i=1}^{g+1} a_i}
     \]
     Using Claim 1, we know \(\prod_{i=1}^g a_i = a_{g+1} - 1\), so:
     \[
     \sum_{i=1}^{g+1} \frac{1}{a_i} = \frac{(a_{g+1} - 1) a_{g+1} + 1}{a_{g+1} \prod_{i=1}^g a_i} = \frac{a_{g+1}^2 - a_{g+1} + 1}{a_{g+1} \prod_{i=1}^g a_i}
     \]
     Simplifying:
     \[
     \sum_{i=1}^{g+1} \frac{1}{a_i} = \frac{\prod_{i=1}^{g+1} a_i - 1}{\prod_{i=1}^{g+1} a_i}
     \]
     This completes the induction. \(\blacksquare\)

3. **Conclusion:**
   From these claims, it follows that each term in the sum is less than 1. Therefore, the minimum real number \(L\) such that for every \(k \in \mathbb{N}\):
   \[
   \sum_{i=1}^k \frac{1}{a_i} < L
   \]
   is \(L = 1\).

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