The positive integer $m$ is a multiple of $101$, and the positive integer $n$ is a multiple of $63$. Their sum is $2018$. Find $m - n$.
1. Let \( m = 101x \) and \( n = 63y \), where \( x \) and \( y \) are positive integers.
2. Given that \( m + n = 2018 \), we can substitute the expressions for \( m \) and \( n \):
   \[
   101x + 63y = 2018
   \]
3. We need to find integer solutions for \( x \) and \( y \). To do this, we can solve the equation modulo 101:
   \[
   101x + 63y \equiv 2018 \pmod{101}
   \]
   Since \( 101x \equiv 0 \pmod{101} \), we have:
   \[
   63y \equiv 2018 \pmod{101}
   \]
4. Simplify \( 2018 \mod 101 \):
   \[
   2018 \div 101 = 19 \quad \text{remainder} \quad 99
   \]
   Therefore:
   \[
   2018 \equiv 99 \pmod{101}
   \]
5. Now we have:
   \[
   63y \equiv 99 \pmod{101}
   \]
6. To solve for \( y \), we need the multiplicative inverse of 63 modulo 101. Using the Extended Euclidean Algorithm, we find that the inverse of 63 modulo 101 is 32. Thus:
   \[
   y \equiv 99 \cdot 32 \pmod{101}
   \]
7. Calculate \( 99 \cdot 32 \mod 101 \):
   \[
   99 \cdot 32 = 3168
   \]
   \[
   3168 \div 101 = 31 \quad \text{remainder} \quad 37
   \]
   Therefore:
   \[
   y \equiv 37 \pmod{101}
   \]
8. Substitute \( y = 37 \) back into the original equation to find \( x \):
   \[
   101x + 63 \cdot 37 = 2018
   \]
   \[
   101x + 2331 = 2018
   \]
   \[
   101x = 2018 - 2331
   \]
   \[
   101x = -313
   \]
   Since \( x \) must be a positive integer, we need to adjust \( y \) by adding multiples of 101 to make \( x \) positive. However, this approach shows that the initial solution might have an error. Let's re-evaluate the integer solutions.

9. Re-evaluate the integer solutions for \( x \) and \( y \):
   \[
   101x + 63y = 2018
   \]
   Testing small values, we find:
   \[
   x = 10, y = 16
   \]
   \[
   101 \cdot 10 + 63 \cdot 16 = 1010 + 1008 = 2018
   \]
   This solution works.

10. Calculate \( m - n \):
    \[
    m - n = 101 \cdot 10 - 63 \cdot 16
    \]
    \[
    m - n = 1010 - 1008 = 2
    \]

The final answer is \(\boxed{2}\).