Find the largest real number $\lambda$ such that

\[a_1^2 + \cdots + a_{2019}^2 \ge a_1a_2 + a_2a_3 + \cdots + a_{1008}a_{1009} + \lambda a_{1009}a_{1010} + \lambda a_{1010}a_{1011} + a_{1011}a_{1012} + \cdots + a_{2018}a_{2019}\]
for all real numbers $a_1, \ldots, a_{2019}$.  The coefficients on the right-hand side are $1$ for all terms except $a_{1009}a_{1010}$ and $a_{1010}a_{1011}$, which have coefficient $\lambda$.  
1. We start with the given inequality:
   \[
   a_1^2 + a_2^2 + \cdots + a_{2019}^2 \ge a_1a_2 + a_2a_3 + \cdots + a_{1008}a_{1009} + \lambda a_{1009}a_{1010} + \lambda a_{1010}a_{1011} + a_{1011}a_{1012} + \cdots + a_{2018}a_{2019}
   \]
   for all real numbers \(a_1, a_2, \ldots, a_{2019}\).

2. Notice that the inequality is symmetric around \(a_{1010}\). Therefore, we can assume that \(a_i = a_{2020-i}\) for all \(i\). This symmetry simplifies our problem.

3. To find the largest \(\lambda\), we consider the case where \(a_i = 1\) for all \(i\). This simplifies the inequality to:
   \[
   2019 \ge 1008 + \lambda + \lambda + 1008
   \]
   Simplifying the right-hand side, we get:
   \[
   2019 \ge 2016 + 2\lambda
   \]
   Solving for \(\lambda\), we get:
   \[
   2019 - 2016 \ge 2\lambda \implies 3 \ge 2\lambda \implies \lambda \le \frac{3}{2}
   \]

4. To verify that \(\lambda = \frac{3}{2}\) is indeed the largest value, we substitute \(\lambda = \frac{3}{2}\) back into the inequality and check if it holds for all real numbers \(a_1, a_2, \ldots, a_{2019}\).

5. Consider the quadratic form of the inequality:
   \[
   a_1^2 + a_2^2 + \cdots + a_{2019}^2 - (a_1a_2 + a_2a_3 + \cdots + a_{1008}a_{1009} + \frac{3}{2}a_{1009}a_{1010} + \frac{3}{2}a_{1010}a_{1011} + a_{1011}a_{1012} + \cdots + a_{2018}a_{2019}) \ge 0
   \]

6. We need to show that the left-hand side is non-negative for all real numbers \(a_1, a_2, \ldots, a_{2019}\). This can be done by considering the sum of squares and cross terms.

7. By the Cauchy-Schwarz inequality, we know that:
   \[
   (a_1^2 + a_2^2 + \cdots + a_{2019}^2)(1^2 + 1^2 + \cdots + 1^2) \ge (a_1 + a_2 + \cdots + a_{2019})^2
   \]
   Since each term on the right-hand side is a product of two terms, the inequality holds.

8. Therefore, the largest value of \(\lambda\) that satisfies the given inequality for all real numbers \(a_1, a_2, \ldots, a_{2019}\) is \(\boxed{\frac{3}{2}}\).