For every positive integer $n$, find the maximum power of $2$ that divides the number $$1 + 2019 + 2019^2 + 2019^3 +.. + 2019^{n-1}.$$
1. **Identify the series and apply the geometric series formula:**

   The given series is:
   \[
   1 + 2019 + 2019^2 + 2019^3 + \ldots + 2019^{n-1}
   \]
   This is a geometric series with the first term \(a = 1\) and common ratio \(r = 2019\). The sum of the first \(n\) terms of a geometric series is given by:
   \[
   S_n = \frac{r^n - 1}{r - 1}
   \]
   Substituting \(r = 2019\), we get:
   \[
   S_n = \frac{2019^n - 1}{2019 - 1} = \frac{2019^n - 1}{2018}
   \]

2. **Analyze the expression for even \(n\):**

   When \(n\) is even, we use the Lifting The Exponent (LTE) lemma to find the highest power of 2 dividing \(2019^n - 1\). The LTE lemma states:
   \[
   v_2(x^n - y^n) = v_2(x - y) + v_2(x + y) + v_2(n) - 1
   \]
   for \(x\) and \(y\) odd integers and \(n\) even. Here, \(x = 2019\) and \(y = 1\), so:
   \[
   v_2(2019^n - 1) = v_2(2019 - 1) + v_2(2019 + 1) + v_2(n) - 1
   \]
   Calculate each term:
   \[
   v_2(2019 - 1) = v_2(2018) = v_2(2 \times 1009) = 1
   \]
   \[
   v_2(2019 + 1) = v_2(2020) = v_2(2^2 \times 505) = 2
   \]
   \[
   v_2(n) = v_2(n)
   \]
   Therefore:
   \[
   v_2(2019^n - 1) = 1 + 2 + v_2(n) - 1 = v_2(n) + 2
   \]
   Since we need the power of 2 dividing \(\frac{2019^n - 1}{2018}\), we subtract \(v_2(2018)\):
   \[
   v_2\left(\frac{2019^n - 1}{2018}\right) = v_2(2019^n - 1) - v_2(2018) = (v_2(n) + 2) - 1 = v_2(n) + 1
   \]

3. **Analyze the expression for odd \(n\):**

   When \(n\) is odd, we consider the expression modulo 2. Since \(2019 \equiv 1 \pmod{2}\), we have:
   \[
   2019^n \equiv 1^n \equiv 1 \pmod{2}
   \]
   Thus:
   \[
   2019^n - 1 \equiv 1 - 1 \equiv 0 \pmod{2}
   \]
   However, since \(2018 \equiv 0 \pmod{2}\), the fraction \(\frac{2019^n - 1}{2018}\) is an integer, but it is not divisible by any higher power of 2. Therefore, the power of 2 dividing the expression is 0.

Conclusion:
- If \(n\) is even, the maximum power of 2 that divides the expression is \(v_2(n) + 1\).
- If \(n\) is odd, the maximum power of 2 that divides the expression is 0.

The final answer is \(\boxed{0}\) if \(n\) is odd, and \(\boxed{v_2(n) + 1}\) if \(n\) is even.