7. (10 points) Fill in each box with a digit so that the multiplication vertical format holds. Then find the sum of the two multiplicands in the equation.

保留源文本的换行和格式，这里的翻译结果为：

7. (10 points) Fill in each box with a digit so that the multiplication vertical format holds. Then find the sum of the two multiplicands in the equation.
Is $\qquad$ .
[Analysis] First, since the tens digit of the second multiplier must be greater than the hundreds digit, the "1" given in the problem cannot be the result of $9 \times 9$. Since multiplying a number in the 200s by a number in the 100s cannot result in a six-digit number, the "1" given in the problem cannot be the result of two 1s multiplied together. Therefore, it must be the units digit of $3 \times 7$. Following this rule, we can reason step by step.

[Solution] Since the tens digit of the second multiplier must be greater than the hundreds digit, the "1" given in the problem cannot be the result of $9 \times 9$. Since multiplying a number in the 200s by a number in the 100s cannot result in a six-digit number, the "1" given in the problem cannot be the result of two 1s multiplied together. Therefore, it must be the units digit of $3 \times 7$.

The "2" given in the problem can only be the result of $1 \square \square \times \square \square 2$ or $2 \square \square \times 1 \square \square$. Combining the above analysis, we get the following four scenarios:
(1) 1 $\square$ $73 \times 7$ $\square 2$
(2) 1 $\square$ $7 \times 3$ $\square$ 2
(3) 2 $\square$ $3 \times 7$ $\square$ $\square 1$
(4) $2 \square 7 \times 3 \square$ $\square$ 1

If it is scenario (1), then the $\square$ in $7 \square 2$ must be greater than 7, which can only be 8 or 9. Analyzing the 0 in the second row, we get $113 \times 782, 163 \times 782$, but neither is valid. If it is scenario (2), the total product cannot be a six-digit number. If it is scenario (3), the row containing the "1" given in the problem is a four-digit number. If it is scenario (4), we try each possibility for the $\square$ in $3 \square 1$ to see if $2 \square 7 \times \square = \square \square 0 \square$, yielding $277 \times 341, 217 \times 361, 267 \times 361, 287 \times 371, 267 \times 391$.

Upon verification, only $287 \times 371 = 106477$ is valid.
Therefore, the answer is: $287 + 371 = 658$.