If [tex]x[/tex] is a number that is both divisible by 4 and 5, then
[tex]\begin{cases}x\equiv0\pmod4\\x\equiv0\pmod5\end{cases}[/tex]
4 and 5 are coprime, so we can use the Chinese remainder theorem to solve this system and find that [tex]x=20n[/tex] is a solution to the system, where [tex]n[/tex] is any integer. Simply put, any multiple of 20 fits the bill.
Now, there are 11 numbers between 100 and 300 that are divisible by 20 (100, 120, 140, and so on). We have [tex]20n=100[/tex] when [tex]n=5[/tex], so the sum we want to compute is
[tex]\displaystyle\sum_{n=5}^{15}20n=\boxed{2200}[/tex]
