Solution

Answer: \(n=k+4\).

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 \(A_{4}, A_{5}, \cdots, A_{l+2}\).