How many integers divide either $2018$ or $2019$? Note: $673$ and $1009$ are both prime.

[i]2019 CCA Math Bonanza Lightning Round #1.1[/i]
To determine how many integers divide either \(2018\) or \(2019\), we need to find the number of divisors of each number and then account for any overlap (common divisors).

1. **Find the prime factorizations:**
   - \(2018 = 2 \times 1009\)
   - \(2019 = 3 \times 673\)

2. **Determine the number of divisors:**
   - For \(2018\), the prime factorization is \(2^1 \times 1009^1\). The number of divisors is \((1+1)(1+1) = 2 \times 2 = 4\).
   - For \(2019\), the prime factorization is \(3^1 \times 673^1\). The number of divisors is \((1+1)(1+1) = 2 \times 2 = 4\).

3. **Identify common divisors:**
   - Since \(2018\) and \(2019\) are coprime (they have no common prime factors), their only common divisor is \(1\).

4. **Apply the principle of inclusion-exclusion:**
   - The total number of divisors of either \(2018\) or \(2019\) is given by:
     \[
     \text{Number of divisors of } 2018 + \text{Number of divisors of } 2019 - \text{Number of common divisors}
     \]
     \[
     = 4 + 4 - 1 = 7
     \]

Therefore, the number of integers that divide either \(2018\) or \(2019\) is \(7\).

The final answer is \(\boxed{7}\)