To solve the expression \(2018^{2020} + 2020^{2018}\), you can consider modulus properties.

The expression simplifies as both large powers yield large numbers, but calculating directly is impractical without a computer. However, this approach examines simplicity by modulus:

Evaluate the expression modulo smaller numbers to infer patterns or behavior:

1. **Modulo 2019:**

- By Fermat's Little Theorem, for any integer \(a\) and prime \(p\), \(a^{p-1} \equiv 1 \pmod{p}\) for \(p\) that doesn't divide \(a\).
- \(2018 \equiv -1 \pmod{2019}\), so \(2018^{2020} \equiv (-1)^{2020} \equiv 1 \pmod{2019}\).
- \(2020 \equiv 1 \pmod{2019}\), so \(2020^{2018} \equiv 1^{2018} \equiv 1 \pmod{2019}\).

Adding them:
\[
2018^{2020} + 2020^{2018} \equiv 1 + 1 \equiv 2 \pmod{2019}
\]

Therefore, the result of this expression is \(\boxed{2}\) when divided by 2019.