For any set $S$ of integers, let $f(S)$ denote the number of integers $k$ with $0 \le k < 2019$ such that there exist $s_1, s_2 \in S$ satisfying $s_1 - s_2 = k$. For any positive integer $m$, let $x_m$ be the minimum possible value of $f(S_1) + \dots + f(S_m)$ where $S_1, \dots, S_m$ are nonempty sets partitioning the positive integers. Let $M$ be the minimum of $x_1, x_2, \dots$, and let $N$ be the number of positive integers $m$ such that $x_m = M$. Compute $100M + N$.

[i]Proposed by Ankan Bhattacharya[/i]
1. **Understanding the function \( f(S) \)**:
   - For a set \( S \) of integers, \( f(S) \) counts the number of integers \( k \) in the range \( 0 \le k < 2019 \) such that there exist \( s_1, s_2 \in S \) with \( s_1 - s_2 = k \).
   - This means \( f(S) \) essentially counts the distinct differences \( s_1 - s_2 \) modulo 2019.

2. **Partitioning the positive integers**:
   - We need to partition the positive integers into \( m \) nonempty sets \( S_1, S_2, \ldots, S_m \) and minimize the sum \( f(S_1) + f(S_2) + \cdots + f(S_m) \).

3. **Lower bound of \( x_m \)**:
   - Consider the numbers \( 1, 2, \ldots, 2019 \). If we partition these into \( m \) sets, each set \( S_i \) will have \( a_i \) elements.
   - The function \( f(S_i) \) will be at least \( a_i \) because the differences \( s_1 - s_2 \) for \( s_1, s_2 \in S_i \) will cover at least \( a_i \) distinct values.
   - Therefore, \( f(S_1) + f(S_2) + \cdots + f(S_m) \geq a_1 + a_2 + \cdots + a_m = 2019 \).

4. **Achieving the lower bound**:
   - To achieve \( f(S_1) + f(S_2) + \cdots + f(S_m) = 2019 \), each set \( S_i \) must cover exactly \( a_i \) distinct differences.
   - This can be achieved if each \( S_i \) is an arithmetic sequence with a common difference that divides 2019.

5. **Possible values of \( r_i \)**:
   - The divisors of 2019 are 1, 3, 673, and 2019.
   - If \( S_i \) is an arithmetic sequence with common difference \( r_i \), then \( r_i \) must be one of these divisors.

6. **Covering system**:
   - We need to find sets \( S_i \) such that the sum of their reciprocals equals 1: \( \sum \frac{1}{r_i} = 1 \).
   - Additionally, \( \sum \lceil \frac{2019}{r_i} \rceil = 2019 \).

7. **Possible partitions**:
   - \( r_i = 1 \): This gives one set covering all integers.
   - \( r_i = 3 \): We need three sets, each covering every third integer.
   - \( r_i = 673 \): We need 673 sets, each covering every 673rd integer.
   - \( r_i = 2019 \): We need 2019 sets, each covering every 2019th integer.

8. **Counting the number of solutions**:
   - The solutions are of the form \( 2019 - 672n \) (for \( n \leq 3 \)), \( 2019 - 2n \) (for \( n \leq 673 \)), and 1.
   - This gives 676 solutions.

The final answer is \( \boxed{ 100M + N = 100 \times 2019 + 676 = 201900 + 676 = 202576 } \).