1. **Stating the problem:** We are given a puzzle where four boxes on the left side are added to four boxes on the right side, resulting in eight boxes in total. The challenge is to fill these boxes with digits 1 to 9, each used exactly once, to satisfy the addition. 2. **Understanding the puzzle:** The puzzle looks like this: $$\boxed{\ }\boxed{\ } + \boxed{\ }\boxed{\ } = \boxed{\ }\boxed{\ }\boxed{\ }\boxed{\ }$$ We need to find digits 1 through 9, each used once, to fill these boxes so that the sum is correct. 3. **Key rules:** - Each digit 1 to 9 is used exactly once. - The sum of the two two-digit numbers equals the four-digit number. 4. **Setting variables:** Let the first two-digit number be $AB$ (digits $A$ and $B$), the second two-digit number be $CD$ (digits $C$ and $D$), and the four-digit sum be $EFGH$ (digits $E$, $F$, $G$, $H$). 5. **Formulating the equation:** $$10A + B + 10C + D = 1000E + 100F + 10G + H$$ 6. **Constraints:** - $A, B, C, D, E, F, G, H$ are digits from 1 to 9. - All digits are distinct. - The sum of two two-digit numbers results in a four-digit number, so the sum must be at least 1000. 7. **Logical deductions:** - The maximum sum of two two-digit numbers is $99 + 98 = 197$, which is less than 1000. - Therefore, the puzzle likely means the sum of two two-digit numbers equals a four-digit number formed by concatenating the digits in the boxes, not a numerical sum. 8. **Alternative interpretation:** The puzzle might be a cryptarithm where the boxes represent digits, and the sum is a concatenation or a different operation. 9. **Conclusion:** Without additional context or rules, the problem as stated cannot be solved as a standard addition because two two-digit numbers cannot sum to a four-digit number. **Final answer:** The problem requires clarification or additional rules to solve correctly.