Let $M$ be the set of the positive rational numbers less than 1, which can be expressed with a 10-distinct digits period in decimal representation.

a) Find the arithmetic mean of all the elements in $M$;

b) Prove that there exists a positive integer $n$, $1<n<10^{10}$, such that $n\cdot a - a$ is a non-negative integer, for all $a\in M$.
### Part (a)

1. **Understanding the Set \( M \)**:
   - The set \( M \) consists of positive rational numbers less than 1, which have a 10-digit repeating decimal period. 
   - Such numbers can be written in the form \( \frac{p}{q} \) where \( q \) is a divisor of \( 10^{10} - 1 \).

2. **Complementary Pairs**:
   - Each element \( a \in M \) has a unique complement \( 1 - a \) such that \( a + (1 - a) = 0.9999999999 \).
   - This is because \( 0.9999999999 \) is equivalent to 1 in decimal representation.

3. **Arithmetic Mean Calculation**:
   - To find the arithmetic mean of all elements in \( M \), consider the sum of each element and its complement.
   - The sum of each pair \( (a, 1 - a) \) is \( 0.9999999999 \).
   - The average of each pair is \( \frac{0.9999999999}{2} = 0.49999999995 \).

4. **Conclusion**:
   - Since every element in \( M \) can be paired with its complement, the arithmetic mean of all elements in \( M \) is \( 0.49999999995 \).

### Part (b)

1. **Understanding the Problem**:
   - We need to find a positive integer \( n \), \( 1 < n < 10^{10} \), such that \( n \cdot a - a \) is a non-negative integer for all \( a \in M \).

2. **Form of \( a \)**:
   - Each \( a \in M \) can be written as \( a = \frac{p}{q} \) where \( q \) is a divisor of \( 10^{10} - 1 \).

3. **Expression Simplification**:
   - Consider \( n \cdot a - a = a(n - 1) \).
   - For \( a(n - 1) \) to be an integer, \( n - 1 \) must be a multiple of the denominator \( q \) of \( a \).

4. **Choosing \( n \)**:
   - Since \( q \) is a divisor of \( 10^{10} - 1 \), we can choose \( n \) such that \( n - 1 \) is a multiple of \( 10^{10} - 1 \).
   - Let \( n = 10^{10} \). Then \( n - 1 = 10^{10} - 1 \), which is a multiple of any \( q \) that divides \( 10^{10} - 1 \).

5. **Verification**:
   - For any \( a = \frac{p}{q} \in M \), \( a(n - 1) = \frac{p}{q} \cdot (10^{10} - 1) \) is an integer because \( q \) divides \( 10^{10} - 1 \).

6. **Conclusion**:
   - Therefore, \( n = 10^{10} \) satisfies the condition \( 1 < n < 10^{10} \) and \( n \cdot a - a \) is a non-negative integer for all \( a \in M \).

The final answer is \( \boxed{ 0.49999999995 } \) for part (a) and \( n = 10^{10} \) for part (b).