The sum of this expression can be computed using hockey stick identity. $$\begin{align*} &1\times 2 + 2\times 3 + 3\times 4 + \cdots + 2018\times 2019 \\ \\=\ &2!\times\left(\frac{1\times 2}{2!} + \frac{2\times 3}{2!} + \frac{3\times 4}{2!} + \cdots + \frac{2018\times 2019}{2!} \right)\\ \\=\ & 2!\times\left(\binom{2}{2}+\binom{3}{2}+\binom{4}{2}+\cdots +\binom{2019}{2}\right)\\ \\=\ & 2!\times \binom{2020}{3}\\ \\=\ & 2\times\frac{2020\times 2019\times 2018}{1\times 2\times 3}\\ \\=\ & 2020\times 673\times 2018\end{align*}$$

Therefore, the answer is $\boxed{0}$.