How can this be possible? Let $a_1, a_2$ and $a_3$ be on a same strip. Then $a_1 = a_2 + a_i + a_m$ which gives $a_1 > a_2$ Implying this we get $a_2 = a_3 + a_n + a_p$ which gives $a_2 > a_3$ then how can $a_3$ be equal to one of $a_1$ and $a_2$ plus something else when it is smaller than both of them and integers are positive

Ferum_2710 wrote: How can this be possible? Let $a_1, a_2$ and $a_3$ be on a same strip. Then $a_1 = a_2 + a_i + a_m$ which gives $a_1 > a_2$ Implying this we get $a_2 = a_3 + a_n + a_p$ which gives $a_2 > a_3$ then how can $a_3$ be equal to one of $a_1$ and $a_2$ plus something else when it is smaller than both of them and integers are positive Any ideas?

I am not sure if I translated it correctly but you can find the romanian wording here: https://ssmr.ro/onm2023.

But do you agree with my idea?

If all the numbers are equal to 1, there is nothing to prove. If the cube contains numbers greater than 1, we assume that there are even numbers written in the cube, and let n be the smallest of them. Then n must be the sum of 3 odd numbers, which is impossible, as the sum of three odd numbers is an odd number. Therefore, all numbers are odd. Assume that there is a strip that does not contain the number 1. Let a be the smallest number on that strip. Since a > 1, it is the sum of three numbers smaller than it and greater than 1, contradicting the minimality of a. Therefore, each strip contains at least one 1, so the cube contains at least 9 of them. This means that there are at most 18 other numbers in the cube, and since all of them are odd and greater than 1, they must be at least 3, 5, 7, ..., 35. Therefore, there are at most 16 numbers in the cube that are less than or equal to 60. Suppose $a_1\le a_2\le \ldots \le a_{18}$ are 18 numbers from a cube. If $a_1>3$, then one of the three numbers in the cube whose sum is $a_1$ must be greater than 1 and less than $a_1$, which contradicts the minimality of $a_1$. Thus, $a_1$ is in ${1,3}$. Therefore, $a_2$ is in ${1,3,5}$. If $a_2\le 3$, then $a_3\le 1+a_1+a_2\le 7$. If $a_2=5$, then $a_1$ and $a_2$ are on the same face of the cube, so they cannot both appear in the expression for $a_3$, which means $a_3\le 1+1+a_2=7$. Thus, $a_3$ is in ${1,3,5,7}$. If $a_3=7$, then $a_3$ is on the same face as $a_2$, so $a_2$ and $a_3$ cannot both appear in the expression for $a_4$, which means $a_4\le a_3+a_1+1\le 7+3+1=11$. If $a_3\le 5$, then if $a_2=5$, then $a_1$ and $a_2$ are on the same face, so $a_4\le a_3+a_2+a_1\le 5+5+1=11$, and if $a_2\le 3$, then $a_4\le a_3+a_2+a_1\le 5+3+3=11$. In any case, we have $a_4\le 11$. In general, we have $a_{k+3}\le a_{k+2}+a_{k+1}+a_k$. If $a_{k+1}$ and $a_{k+2}$ are on the same face, then they cannot both appear in the expression for $a_{k+3}$. Thus, we have $a_{k+3}\le a_{k+2}+a_{k+1}+a_{k}\le a_{k+1}+2a_{k}+2a_{k-1}$. If $a_{k+1}$ and $a_{k+2}$ are on different faces, then $a_{k+3}\le a_{k+2}+a_{k+1}+a_k\le a_{k+1}+2a_k+a_{k-1}+a_{k-2}\le a_{k+1}+2a_k+2a_{k-1}$. Thus, $a_{k+3}\le a_{k+1}+2a_k+2a_{k-1}$ for any $k\ge 2$. From this inequality, we can successively deduce that $a_5\le 23$, $a_6\le 35$, and $a_7\le 59$, so at most 16 numbers are less than or equal to 60.