International Mathematical Olympiad

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 $k = 2l$ 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 $k+4 = 2l+4$ 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 give a construction for odd $k = 2l+1$ with $l \geqslant 1$ (since $k \geqslant 2$). To that end, we modify the construction for $k = 2l$ 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 $A_{4}, A_{5}, \cdots, A_{l+2}$.