Problem (Algebra):
Let $k \geqslant 2$ be an integer. Find the smallest integer $n \geqslant k+1$ with the property that there exists a set of $n$ distinct real numbers such that each of its elements can be written as a sum of $k$ other distinct elements of the set.

Solution:
First we show that $n \geqslant k+4$. Suppose that there exists such a set with $n$ numbers and denote them by $a_{1}<a_{2}<\cdots<a_{n}$.

Note that in order to express $a_{1}$ as a sum of $k$ distinct elements of the set, we must have $a_{1} \geqslant a_{2}+\cdots+a_{k+1}$ and, similarly for $a_{n}$, we must have $a_{n-k}+\cdots+a_{n-1} \geqslant a_{n}$. We also know that $n \geqslant k+1$.

If $n=k+1$ then we have $a_{1} \geqslant a_{2}+\cdots+a_{k+1}>a_{1}+\cdots+a_{k} \geqslant a_{k+1}$, which gives a contradiction.

If $n=k+2$ then we have $a_{1} \geqslant a_{2}+\cdots+a_{k+1} \geqslant a_{k+2}$, that again gives a contradiction.

If $n=k+3$ then we have $a_{1} \geqslant a_{2}+\cdots+a_{k+1}$ and $a_{3}+\cdots+a_{k+2} \geqslant a_{k+3}$. Adding the two inequalities we get $a_{1}+a_{k+2} \geqslant a_{2}+a_{k+3}$, again a contradiction.

It remains to give an example of a set with $k+4$ elements satisfying the condition of the problem. We start with the case when $k=2 l$ and $l \geqslant 1$. In that case, denote by $A_{i}=\{-i, i\}$ and take the set $A_{1} \cup \cdots \cup A_{l+2}$, which has exactly $k+4=2 l+4$ elements. We are left to show that this set satisfies the required condition.

Note that if a number $i$ can be expressed in the desired way, then so can $-i$ by negating the expression. Therefore, we consider only $1 \leqslant i \leqslant l+2$.

If $i<l+2$, we sum the numbers from some $l-1$ sets $A_{j}$ with $j \neq 1, i+1$, and the numbers $i+1$ and -1 .

For $i=l+2$, we sum the numbers from some $l-1$ sets $A_{j}$ with $j \neq 1, l+1$, and the numbers $l+1$ and 1 .

It remains to a give a construction for odd $k=2 l+1$ with $l \geqslant 1$ (since $k \geqslant 2$ ). To that end, we modify the construction for $k=2 l$ by adding 0 to the previous set.

This is a valid set as 0 can be added to each constructed expression, and 0 can be expressed as follows: take the numbers $1,2,-3$ and all the numbers from the remaining $l-1$ sets $A_{4}, A_{5}, \cdots, A_{l+2}$.

Final answer: $n=k+4$