11.5. Find all natural $n$ for which all natural numbers from 1 to $n$ inclusive can be arranged in a row in such an order that the sum of the first $k$ numbers from the left will either divide the sum of all $n-k$ remaining numbers or be divisible by it for any $k$ from 1 to $n-1$.
Answer: $n=3,4,5$.

Solution. Examples for these numbers: $\{1,2,3\},\{1,4,3,2\},\{1,4,5,2,3\}$.

Let $a_{1}, a_{2}, \ldots, a_{n}$ be all numbers from 1 to $n$, written in the required order. Denote their sum by $S$, and consider the maximum $k$ such that $A_{k}=a_{1}+a_{2}+\ldots+a_{k}$ does not exceed $\frac{S}{2}$.

a) $A_{k}<\frac{S}{2}$. Then the numbers $A_{k}$ and $B_{k}=a_{k+2}+\ldots+a_{n}=S-A_{k}-a_{k+1}$ are proper divisors of the number $S$, less than $\frac{S}{2}$, so both are no more than $\frac{S}{3}$. Therefore, $a_{k+1}=S-A_{k}-B_{k} \geq S-2 \frac{S}{3}=\frac{S}{3}$, from which $\frac{S}{3}=\frac{n(n+1)}{6} \leq a_{k+1} \leq n$ and $n \leq 5$.

b) $A_{k}=\frac{S}{2}$. Then the numbers $A_{k-1}=a_{1}+a_{2}+\ldots+a_{k-1}=\frac{S}{2}-a_{k}$ and $B_{k}=a_{k+2}+\ldots+a_{n}=\frac{S}{2}-a_{k+1}$
are distinct proper divisors of the number $S$, less than $\frac{S}{2}$, so one of them is no more than $\frac{S}{4}$, we can assume $A_{k-1} \leq \frac{S}{4}$. Then $a_{k}=\frac{S}{2}-A_{k-1} \geq \frac{S}{4}$, from which $\frac{S}{4}=\frac{n(n+1)}{8} \leq a_{k} \leq n$ and $n \leq 7$. Moreover, for $n=6$ case b) is impossible due to the oddness of $S$. For $n=7$ the number $S$ is not divisible by 3, so one of the divisors $A_{k-1}$ and $B_{k}$ does not exceed $\frac{S}{5}$, which allows us to improve the estimate: $\frac{S}{2}-\frac{S}{5}=\frac{3 S}{10}=\frac{3 n(n+1)}{20} \leq a_{k} \leq n$, from which $n \leq \frac{20}{3}-1=\frac{17}{3}$ and $n \leq 5$.

Grading criteria: Finding examples for all $n=3,4,5: 1$ point. Estimating $n \leq 7: 3$ points. Estimating $n \leq 5: 6$ points.