For a positive integer $n$, $S_n$ is the set of positive integer $n$-tuples $(a_1,a_2, \cdots ,a_n)$ which satisfies the following.

(i). $a_1=1$.

(ii). $a_{i+1} \le a_i+1$.

For $k \le n$, define $N_k$ as the number of $n$-tuples $(a_1, a_2, \cdots a_n) \in S_n$ such that $a_k=1, a_{k+1}=2$.

Find the sum $N_1 + N_2+ \cdots N_{k-1}$.
1. **Define the transformation and constraints:**
   We start by defining \( b_i = i - a_i \). Given the constraints:
   - \( a_1 = 1 \) implies \( b_1 = 1 - 1 = 0 \).
   - \( a_{i+1} \le a_i + 1 \) implies \( b_{i+1} \ge b_i \).
   - Since \( a_i \) is a positive integer, \( 1 \le a_i \le i \), thus \( 0 \le b_i \le i - 1 \).

2. **Interpret the constraints in terms of \( b_i \):**
   The sequence \( \{b_i\} \) is non-decreasing and each \( b_i \) is bounded by \( i - 1 \). This sequence can be visualized as a walk on a grid where each step is either horizontal or vertical, and the walk never goes below the line \( y = 0 \).

3. **Catalan-like walk:**
   The sequence \( \{b_i\} \) corresponds to a Catalan-like walk. Specifically, the number of such walks from \((0,0)\) to \((n, b_n)\) that stay above the line \( y = 0 \) is given by the Catalan numbers.

4. **Condition for \( N_k \):**
   The condition \( a_k = 1 \) and \( a_{k+1} = 2 \) translates to \( b_k = k - 1 \) and \( b_{k+1} = k - 1 \). This means that at step \( k \), the walk reaches the line \( y = k - 1 \) and stays there for the next step.

5. **Counting the walks:**
   The number of such walks can be decomposed into:
   - The number of walks from \((0,0)\) to \((k-1, k-1)\), which is given by the Catalan number \( C_{k-1} \).
   - The number of walks from \((k, k-1)\) to \((n, b_n)\) that stay above the line \( y = k - 1 \). This is given by \( C_{n-k+1} - C_{n-k} \), where \( C_{n-k+1} \) counts all walks from \((k, k-1)\) to \((n, n-k)\) and \( C_{n-k} \) subtracts those that touch the line \( y = k - 2 \).

6. **Summing up \( N_k \):**
   Therefore, \( N_k = C_{k-1} (C_{n-k+1} - C_{n-k}) \).

7. **Summing \( N_1 + N_2 + \cdots + N_{n-1} \):**
   We need to sum \( N_k \) from \( k = 1 \) to \( k = n-1 \):
   \[
   \sum_{k=1}^{n-1} N_k = \sum_{k=1}^{n-1} C_{k-1} (C_{n-k+1} - C_{n-k})
   \]

8. **Using the Catalan recurrence relation:**
   The Catalan numbers satisfy the recurrence relation \( C_{n+1} = \sum_{i=0}^{n} C_i C_{n-i} \). Using this, we can simplify the sum:
   \[
   \sum_{k=1}^{n-1} C_{k-1} (C_{n-k+1} - C_{n-k}) = C_{n+1} - 2C_n
   \]

Conclusion:
\[
\boxed{C_{n+1} - 2C_n}
\]