4. Arrange all positive odd numbers in ascending order, take the first number as $a_{1}$, take the sum of the next two numbers as $a_{2}$, then take the sum of the next three numbers as $a_{3}$, and so on, to get the sequence $\left\{a_{n}\right\}$, that is,
$$
a_{1}=1, a_{2}=3+5, a_{3}=7+9+11, \cdots \cdots
$$

Then $a_{1}+a_{2}+\cdots+a_{20}=$ . $\qquad$
4. 44100 .

Notice that, $\sum_{i=1}^{20} a_{i}$ is the sum of the first $\sum_{i=1}^{20} i=210$ odd numbers. Therefore, $210^{2}=44100$.