How many integers can be expressed in the form: $\pm 1 \pm 2 \pm 3 \pm 4 \pm \cdots \pm 2018$?
1. **Determine the minimum and maximum values:**
   The minimum value is obtained by taking the negative of all the integers from 1 to 2018:
   \[
   -1 - 2 - 3 - \cdots - 2018 = -\sum_{k=1}^{2018} k = -\frac{2018 \cdot 2019}{2} = -2037171
   \]
   The maximum value is obtained by taking the positive of all the integers from 1 to 2018:
   \[
   1 + 2 + 3 + \cdots + 2018 = \sum_{k=1}^{2018} k = \frac{2018 \cdot 2019}{2} = 2037171
   \]

2. **Determine the parity of the possible values:**
   Each integer in the form $\pm 1 \pm 2 \pm 3 \pm \cdots \pm 2018$ can be expressed as a sum of the integers from 1 to 2018 with some of them negated. The sum of the integers from 1 to 2018 is:
   \[
   \sum_{k=1}^{2018} k = \frac{2018 \cdot 2019}{2} = 2037171
   \]
   Since 2037171 is odd, any sum obtained by changing the signs of some of these integers will also be odd. This is because changing the sign of an even number of terms does not change the parity of the sum, and changing the sign of an odd number of terms changes the parity of the sum, but since the total sum is odd, the result will always be odd.

3. **Count the number of odd integers between the minimum and maximum values:**
   The range of values is from $-2037171$ to $2037171$. To find the number of odd integers in this range, note that the sequence of odd integers between $-2037171$ and $2037171$ is:
   \[
   -2037171, -2037169, -2037167, \ldots, 2037167, 2037169, 2037171
   \]
   This sequence is an arithmetic sequence with the first term $a = -2037171$, the common difference $d = 2$, and the last term $l = 2037171$. The number of terms $n$ in this sequence can be found using the formula for the $n$-th term of an arithmetic sequence:
   \[
   l = a + (n-1)d
   \]
   Substituting the values, we get:
   \[
   2037171 = -2037171 + (n-1) \cdot 2
   \]
   Solving for $n$:
   \[
   2037171 + 2037171 = (n-1) \cdot 2
   \]
   \[
   4074342 = (n-1) \cdot 2
   \]
   \[
   n-1 = \frac{4074342}{2} = 2037171
   \]
   \[
   n = 2037171 + 1 = 2037172
   \]

Therefore, the number of integers that can be expressed in the form $\pm 1 \pm 2 \pm 3 \pm \cdots \pm 2018$ is $\boxed{2037172}$.