The arrangements found so far, two having $4$ at the bottom and two having $3$ at the bottom (plus their reflections), are the only solutions.
The key observation is that each of the four rows, from top to bottom, must contain exactly one of the numbers $1, 2, 3$ and $4$. In addition, $10$ in the top row must sit next to the number $1, 2, 3$ or $4$ that is in the top row.
To see that, let $d$ be the number at the bottom. Let $a$ and $b$ be the numbers in the third row (from the top), with $a > b$. Then $d = a - b$. Then let $u$ and $v$ be the numbers in the second row that sit above a, with $u > v$. So $a = u - v$. Finally, let $u = x - y$ with $x, y$ in the top row, $x > y$. So in the end we have $x = d + b + v + y$. $d, b, v, y$ are distinct, positive integer, and their sum is at most $10$; so $x$ must be $10$, and $d, b, v, y$ must be a permutation of $1, 2, 3, 4$.
Now it takes about $10$ minutes to try the various arrangements that satisfy this condition. I won't show them here (no good notation for that), but as you start from the bottom, with $d = 1, 2, 3 or 4$, and you work your way up, most possible arrangements collapse quickly because you will need to use two of the numbers $1, 2, 3, 4$ in the same row (usually the second row from the top) to make the differences work, and that violates the condition we derived above.
This solution requires some manual work but no computer verification.
Perhaps the main lesson would be:
The greedy approach is not always optimal
What would be the greedy method? Pack the truck with heaviest boxes first until you cannot add any more boxes.
This results in a first tour with 81+73+37=191 and a second tour with 67+49+34+30=180 and then a final tour with the one remaining 26 kg.
The fact that the 26 kg just barely fail to fit into the second tour are perhaps intended to annoy you and make you play around to somehow achieve two tours.
- Replacing the 37 of tour with something smaller? There is no way to get four boxes with 81+73 as already 26+30 is too much.
- Replace the 73 with something smaller? doing so and continuing greedily, we arrive at 81+67+49=197 and the rest on the second tour - and done!
Given the age, however, I suppose this problem would be more appropriate if the trial-and-error phase could be performed physically (e.g., moving around suitable paper cutouts). But the numbers used in this problem do not seem to allow that easily.
Best Answer
You can reduce the effort to a level that can be handled without a computer.
Let's write the calculation as
\begin{array}{cc} &A&B\\ \times&&C\\\hline &D&E\\ +&F&G\\\hline &H&I \end{array}
There are strong restrictions on $A$ and $C$, since their product must be single-digit. We can't have $C=1$, since otherwise $DE=AB$. So either $A=1$, or $(A,C)$ must be one of $(2,3)$, $(2,4)$, $(3,2)$ or $(4,2)$. We can exclude $(2,4)$ and $(4,2)$: This implies $D=8$, $F=1$ and $H=9$, but then adding $E$ and $G$ mustn't result in a carry, which is impossible, since $I$ can't be $8$ or $9$ and $E+G$ must be at least $3+5$.
So unless $A=1$, $(A,C)$ must be either $(2,3)$ or $(3,2)$. Also $B\notin\{1,5\}$. That leaves $10$ possible triples for $(A,B,C)$, which each determine values for $D$ and $E$, and it shouldn't be too much effort to go through those $10$ cases, find the $4$ remaining options for $F$, $G$, $H$ and $I$ and show that they don't work out.
That leaves us with $A=1$. In that case the options for $(C,B)$ can be enumerated as $(2,7)$, $(2,8)$, $(2,9)$, $(3,4)$, $(3,6)$, $(3,8)$, $(3,9)$, $(4,3)$, $(4,7)$, $(4,8)$, $(4,9)$, $(6,3)$, $(6,7)$, $(6,9)$, $(7,2)$, $(7,4)$, $(7,6)$, $(7,8)$, $(7,9)$, $(8,2)$, $(8,3)$, $(8,4)$, $(8,7)$, $(8,9)$. That's $24$ more cases, for a total of $34$ cases to be checked.