IMO Shortlist

If there are no $A$-partitions of $n$, the result is vacuously true. Otherwise, let $k_{\text{min}}$ be the minimum number of parts in an $A$-partition of $n$, and let $n=a_{1}+\cdots+a_{k_{\min}}$ be an optimal partition. Denote by $s$ the number of different parts in this partition, so we can write $S=\left\{a_{1}, \ldots, a_{k_{\text{min}}}\right\}=\left\{b_{1}, \ldots, b_{s}\right\}$ for some pairwise different numbers $b_{1}<\cdots<b_{s}$ in $A$.
If $s>\sqrt[3]{6 n}$, we will prove that there exist subsets $X$ and $Y$ of $S$ such that $|X|<|Y|$ and $\sum_{x \in X} x=\sum_{y \in Y} y$. Then, deleting the elements of $Y$ from our partition and adding the elements of $X$ to it, we obtain an $A$-partition of $n$ into less than $k_{\text{min}}$ parts, which is the desired contradiction.
For each positive integer $k \leqslant s$, we consider the $k$-element subset
$$S_{1,0}^{k}:=\left\{b_{1}, \ldots, b_{k}\right\}$$
as well as the following $k$-element subsets $S_{i, j}^{k}$ of $S$ :
$$S_{i, j}^{k}:=\left\{b_{1}, \ldots, b_{k-i}, b_{k-i+j+1}, b_{s-i+2}, \ldots, b_{s}\right\}, \quad i=1, \ldots, k, \quad j=1, \ldots, s-k$$
Pictorially, if we represent the elements of $S$ by a sequence of dots in increasing order, and represent a subset of $S$ by shading in the appropriate dots, we have:

Denote by $\Sigma_{i, j}^{k}$ the sum of elements in $S_{i, j}^{k}$. Clearly, $\Sigma_{1,0}^{k}$ is the minimum sum of a $k$-element subset of $S$. Next, for all appropriate indices $i$ and $j$ we have
$$\Sigma_{i, j}^{k}=\Sigma_{i, j+1}^{k}+b_{k-i+j+1}-b_{k-i+j+2}<\Sigma_{i, j+1}^{k} \quad \text{ and } \quad \Sigma_{i, s-k}^{k}=\Sigma_{i+1,1}^{k}+b_{k-i}-b_{k-i+1}<\Sigma_{i+1,1}^{k} .$$
Therefore
$$1 \leqslant \Sigma_{1,0}^{k}<\Sigma_{1,1}^{k}<\Sigma_{1,2}^{k}<\cdots<\Sigma_{1, s-k}^{k}<\Sigma_{2,1}^{k}<\cdots<\Sigma_{2, s-k}^{k}<\Sigma_{3,1}^{k}<\cdots<\Sigma_{k, s-k}^{k} \leqslant n .$$
To see this in the picture, we start with the $k$ leftmost points marked. At each step, we look for the rightmost point which can move to the right, and move it one unit to the right. We continue until the $k$ rightmost points are marked. As we do this, the corresponding sums clearly increase.
For each $k$ we have found $k(s-k)+1$ different integers of the form $\sum_{i, j}^{k}$ between 1 and $n$. As we vary $k$, the total number of integers we are considering is
$$\sum_{k=1}^{s}(k(s-k)+1)=s \cdot \frac{s(s+1)}{2}-\frac{s(s+1)(2 s+1)}{6}+s=\frac{s\left(s^{2}+5\right)}{6}>\frac{s^{3}}{6}>n .$$
Since they are between 1 and $n$, at least two of these integers are equal. Consequently, there exist $1 \leqslant k<k^{\prime} \leqslant s$ and $X=S_{i, j}^{k}$ as well as $Y=S_{i^{\prime}, j^{\prime}}^{k^{\prime}}$ such that
$$\sum_{x \in X} x=\sum_{y \in Y} y, \quad \text{ but } \quad|X|=k<k^{\prime}=|Y|$$
as required. The result follows.

Assume, to the contrary, that the statement is false, and choose the minimum number $n$ for which it fails. So there exists a set $A \subseteq\{1, \ldots, n\}$ together with an optimal $A$ partition $n=a_{1}+\cdots+a_{k_{\text{min}}}$ of $n$ refuting our statement, where, of course, $k_{\text{min}}$ is the minimum number of parts in an $A$-partition of $n$. Again, we define $S=\left\{a_{1}, \ldots, a_{k_{\text{min}}}\right\}=\left\{b_{1}, \ldots, b_{s}\right\}$ with $b_{1}<\cdots<b_{s}$; by our assumption we have $s>\sqrt[3]{6 n}>1$. Without loss of generality we assume that $a_{k_{\min}}=b_{s}$. Let us distinguish two cases.

Case 1. $b_{s} \geqslant \frac{s(s-1)}{2}+1$.
Consider the partition $n-b_{s}=a_{1}+\cdots+a_{k_{\text{min}}-1}$, which is clearly a minimum $A$-partition of $n-b_{s}$ with at least $s-1 \geqslant 1$ different parts. Now, from $n<\frac{s^{3}}{6}$ we obtain
$$n-b_{s} \leqslant n-\frac{s(s-1)}{2}-1<\frac{s^{3}}{6}-\frac{s(s-1)}{2}-1<\frac{(s-1)^{3}}{6}$$
so $s-1>\sqrt[3]{6\left(n-b_{s}\right)}$, which contradicts the choice of $n$.

Case 2. $b_{s} \leqslant \frac{s(s-1)}{2}$.
Set $b_{0}=0, \Sigma_{0,0}=0$, and $\Sigma_{i, j}=b_{1}+\cdots+b_{i-1}+b_{j}$ for $1 \leqslant i \leqslant j<s$. There are $\frac{s(s-1)}{2}+1>b_{s}$ such sums; so at least two of them, say $\Sigma_{i, j}$ and $\Sigma_{i^{\prime}, j^{\prime}}$, are congruent modulo $b_{s}$ (where $(i, j) \neq\left(i^{\prime}, j^{\prime}\right)$ ). This means that $\Sigma_{i, j}-\Sigma_{i^{\prime}, j^{\prime}}=r b_{s}$ for some integer $r$. Notice that for $i \leqslant j<k<s$ we have
$$0<\Sigma_{i, k}-\Sigma_{i, j}=b_{k}-b_{j}<b_{s},$$
so the indices $i$ and $i^{\prime}$ are distinct, and we may assume that $i>i^{\prime}$. Next, we observe that $\Sigma_{i, j}-\Sigma_{i^{\prime}, j^{\prime}}=\left(b_{i^{\prime}}-b_{j^{\prime}}\right)+b_{j}+b_{i^{\prime}+1}+\cdots+b_{i-1}$ and $b_{i^{\prime}} \leqslant b_{j^{\prime}}$ imply
$$-b_{s}<-b_{j^{\prime}}<\Sigma_{i, j}-\Sigma_{i^{\prime}, j^{\prime}}<\left(i-i^{\prime}\right) b_{s}$$
so $0 \leqslant r \leqslant i-i^{\prime}-1$.
Thus, we may remove the $i$ terms of $\Sigma_{i, j}$ in our $A$-partition, and replace them by the $i^{\prime}$ terms of $\Sigma_{i^{\prime}, j^{\prime}}$ and $r$ terms equal to $b_{s}$, for a total of $r+i^{\prime}<i$ terms. The result is an $A$-partition of $n$ into a smaller number of parts, a contradiction.