Sure, here is the translation:

```
21. Let $A_{n}$ represent the unit digit of the result of $n$ sevens, for example:
$$
A_{1}=7, A_{2}=9, A_{3}=3, \cdots,
$$

then $A_{1}+A_{2}+A_{3}+\cdots+A_{2013}=$ $\qquad$
```
【Analysis】The cycle of the unit digit of the product of several 7s is: 7, 9, 3, 1, with a cycle of four times. For the unit digit of the product of 2013 sevens, we have: $2013 \div 4=503 \cdots 1$, which means there are 503 complete cycles of unit digits, plus the first digit of the first cycle, 7.

【Solution】Solution: The unit digit of $7^{n}$ cycles through $7, 9, 3, 1$,
$$
\begin{array}{l}
2013 \div 4=503 \cdots 1, \\
A_{1}+A_{2}+A_{3}+\cdots+A_{2013}=503 \times(7+9+3+1)+7 \\
=503 \times 20+7, \\
=10060+7, \\
=10067 .
\end{array}
$$

Therefore, the answer is: 10067.
【Comment】This problem examines the issue of unit digits and periodicity.