Find all positive integer $n$, such that there exists $n$ points $P_1,\ldots,P_n$ on the unit circle , satisfying the condition that for any point $M$ on the unit circle, $\sum_{i=1}^n MP_i^k$ is a fixed value for 
\\a) $k=2018$
\\b) $k=2019$.
To solve this problem, we need to find all positive integers \( n \) such that there exist \( n \) points \( P_1, P_2, \ldots, P_n \) on the unit circle satisfying the condition that for any point \( M \) on the unit circle, the sum \( \sum_{i=1}^n MP_i^k \) is a fixed value for \( k = 2018 \) and \( k = 2019 \).

### Part (a): \( k = 2018 \)

1. **Expressing \( MP_i \) in terms of angles:**
   Let \(\text{arg}(M) = x\) and \(\text{arg}(P_i) = y_i\). The distance \( MP_i \) can be expressed as:
   \[
   MP_i = |2 \sin\left(\frac{x - y_i}{2}\right)|
   \]

2. **Expanding \( |2 \sin(x)|^{2018} \):**
   Note that:
   \[
   |2 \sin(x)|^{2018} = (e^{ix} - e^{-ix})^{2018} = \left(2i \sin(x)\right)^{2018}
   \]
   This can be expanded using the binomial theorem:
   \[
   (e^{ix} - e^{-ix})^{2018} = \sum_{j=-1009}^{1009} a_j e^{2ijx}
   \]
   where \( a_j \) are binomial coefficients.

3. **Summing over all points \( P_i \):**
   \[
   f(x) = \sum_{i=1}^n MP_i^{2018} = \sum_{i=1}^n \left(2 \sin\left(\frac{x - y_i}{2}\right)\right)^{2018}
   \]
   This can be written as:
   \[
   f(x) = \sum_{i=1}^n \sum_{j=-1009}^{1009} a_j e^{ij(x - y_i)}
   \]

4. **Condition for \( f(x) \) to be constant:**
   For \( f(x) \) to be constant, the coefficients of all non-zero \( e^{ijx} \) terms must be zero. This implies:
   \[
   \sum_{i=1}^n e^{-ijy_i} = 0 \quad \text{for all } j \neq 0
   \]

5. **Choosing \( y_i \) values:**
   If \( n > 1009 \), we can choose \( y_i = \frac{2\pi (i-1)}{n} \) for \( 1 \leq i \leq n \). This ensures:
   \[
   \sum_{i=1}^n e^{-ijy_i} = \frac{e^{-2\pi ij} - 1}{e^{-\frac{2\pi ij}{n}} - 1} = 0 \quad \text{for } |j| \leq 1009
   \]

6. **Conclusion for \( k = 2018 \):**
   Therefore, for \( k = 2018 \), any \( n > 1009 \) works.

### Part (b): \( k = 2019 \)

1. **Expressing \( MP_i \) in terms of angles:**
   Similar to part (a), let \(\text{arg}(M) = x\) and \(\text{arg}(P_i) = y_i\). The distance \( MP_i \) can be expressed as:
   \[
   MP_i = |2 \sin\left(\frac{x - y_i}{2}\right)|
   \]

2. **Expanding \( |2 \sin(x)|^{2019} \):**
   Note that:
   \[
   |2 \sin(x)|^{2019} = (e^{ix} - e^{-ix})^{2019} = \left(2i \sin(x)\right)^{2019}
   \]
   This can be expanded using the binomial theorem:
   \[
   (e^{ix} - e^{-ix})^{2019} = \sum_{j=-1009}^{1010} a_j e^{(2j-1)ix}
   \]
   where \( a_j \) are binomial coefficients.

3. **Summing over all points \( P_i \):**
   \[
   f(x) = \sum_{i=1}^n MP_i^{2019} = \sum_{i=1}^n \left(2 \sin\left(\frac{x - y_i}{2}\right)\right)^{2019}
   \]
   This can be written as:
   \[
   f(x) = \sum_{i=1}^n \sum_{j=-1009}^{1010} a_j e^{\frac{(2j-1)i(x - y_i)}{2}}
   \]

4. **Condition for \( f(x) \) to be constant:**
   For \( f(x) \) to be constant, the coefficients of all non-zero \( e^{\frac{(2j-1)ix}{2}} \) terms must be zero. This implies:
   \[
   \sum_{i=1}^n e^{\frac{i(x - y_i)}{2}} = 0
   \]

5. **Contradiction:**
   However, each individual \(\text{Im}(e^{\frac{i(x - y_i)}{2}}) = \sin\left(\frac{x - y_i}{2}\right)\) is positive, leading to a contradiction.

6. **Conclusion for \( k = 2019 \):**
   Therefore, for \( k = 2019 \), no \( n \) works.

The final answer is \( \boxed{ n > 1009 } \) for \( k = 2018 \) and no \( n \) for \( k = 2019 \).