Let $M= \{ 1, 2, \cdots, 19 \}$ and $A = \{ a_{1}, a_{2}, \cdots, a_{k}\}\subseteq M$. Find the least $k$ so that for any $b \in M$, there exist $a_{i}, a_{j}\in A$, satisfying $b=a_{i}$ or $b=a_{i}\pm a_{i}$ ($a_{i}$ and $a_{j}$ do not have to be different) .
To solve the problem, we need to find the smallest subset \( A \subseteq M \) such that for any \( b \in M \), there exist \( a_i, a_j \in A \) satisfying \( b = a_i \) or \( b = a_i \pm a_j \). 

1. **Initial Consideration for \( k \leq 3 \)**:
   - If \( k \leq 3 \), then \( A \) has at most three elements. Let these elements be \( a < b < c \).
   - The possible values that can be produced are \( a, b, c, 2a, 2b, 2c, a+b, a+c, b+c, c-a, c-b, b-a \).
   - This gives at most 12 distinct values, which is insufficient to cover all 19 values in \( M \). Hence, \( k \geq 4 \).

2. **Consideration for \( k = 4 \)**:
   - Let \( A = \{a, b, c, d\} \) with \( a < b < c < d \).
   - The possible values that can be produced are \( a, b, c, d, 2a, 2b, 2c, 2d, a+b, a+c, a+d, b+c, b+d, c+d, d-a, d-b, d-c, c-b, c-a, b-a \).
   - This gives at most 20 distinct values. Since we need exactly 19 values, there must be some overlap or an out-of-range value.

3. **Analyzing the Range and Overlap**:
   - Since \( 2d \) should be at least 19 (to cover the largest value), \( d \) must be at least 10.
   - If \( 2d = 19 \), then \( d = 9.5 \), which is not an integer. Hence, \( 2d \neq 19 \).
   - Therefore, \( d + c = 19 \). Given \( c \geq 3 \), we have \( 10 \leq d \leq 16 \) and \( c = 19 - d \).

4. **Checking Possible Values for \( d \)**:
   - For \( d = 10 \), \( c = 9 \), and \( b = 7 \). Trying all possible \( a \) values, no solution is found.
   - For \( d = 11, 12, 13, 14, 15, 16 \) and \( c = 19 - d \), only \( d + b \) can produce 18, so \( b = c - 1 \). Trying all possible \( a \) values, no solution is found.

5. **Considering \( k = 5 \)**:
   - We need to find a set \( A \) with 5 elements that can produce all values from 1 to 19.
   - One such set is \( A = \{2, 6, 7, 8, 11\} \). This set can produce all values from 1 to 19 by combinations of \( a_i \pm a_j \).

6. **Verification**:
   - Another example for \( k = 5 \) is \( A = \{1, 2, 4, 10, 17\} \). This set also covers all values from 1 to 19.

Conclusion:
The least \( k \) such that for any \( b \in M \), there exist \( a_i, a_j \in A \) satisfying \( b = a_i \) or \( b = a_i \pm a_j \) is \( k = 5 \).

The final answer is \( \boxed{5} \).