10. (10 points) In the addition problem below, if different Chinese characters represent different digits, making the equation true, then the maximum value of the four-digit number Huabei Chusai is $\qquad$
$$
\begin{array}{cccc} 
& & \text { Tu } & \text { Nian } \\
\text { Shi } & \begin{array}{c}
\text { Gong } \\
\text { Jie } \\
\text { Hua }
\end{array} & \text { Chu } & \text { Sai } \\
\hline 2 & 0 & 1 & 1
\end{array}
$$
【Analysis】To make the four-digit number "Huabei Chusai" take the maximum value, first determine the value of "Hua", then further determine the values of "Bei", "Chu", and "Sai". According to this, use the casting out nines method to solve the problem.

【Solution】Solution: Observing the problem, it is clear that Hua $=1$. There are a total of 9 digits, so one of the digits from 0 to 9 cannot be used. According to the casting out nines method, 5 cannot be used. Each time a digit is carried over, the sum of the digits decreases by $9,0+1+2+3+4+6+7+8+9-(2+0+1+1)$ $=36$, so there are 4 carries in total, meaning one of the units or tens needs to carry over 2. There are two possibilities:
(1) The sum of the units digits is 11, the sum of the tens digits is 20, and the sum of the hundreds digits is 8;
(2) The sum of the units digits is 21, the sum of the tens digits is 9, and the sum of the hundreds digits is 9;
To make the four-digit number "Huabei Chusai" as large as possible, "Bei" should be as large as possible, and "Chu" should be as small as possible, with "Chu" being the smallest at 2, at which point "Bei" is 7;

The remaining digits are $0, 3, 4, 6, 8, 9$. When the sum of the units is 21, it is clear that $4+8+9=21$; the sum of the tens is 9, then the remaining digits are exactly $0+3+6=9$, so the largest four-digit number is 1769.

Answer: The largest four-digit number is 1769.
Therefore, the answer is: 1769.