https://aops.com/community/p11418512 USJL wrote: For $i=1,\ldots,k$, let $A_i$ be the set of the numbers in the $i$-th deck. Lemma. For any nonempty proper subset $S\subseteq [n]$, there exists $i=1,\ldots,k$ such that $A_i\cup S$ and $A_i\backslash S$ are both nonempty. Proof of Lemma: Suppose that there exists such $S$ that contradicts the lemma. Suppose that $n'=|S|<n$ and there are $k'<k$ indices $i$ such that $A_i\subseteq S$. Then for any number in $|S|$, they must appear $k$ times among the $k'$ decks. On the other hand, for any deck that the corresponding set is contained in $S$, it must contain $n$ cards that have numbers in $S$. Therefore $n'k=nk'$, which is a contradiction because $\gcd(k,n)=1$. Back to the problem. We are going to claim that there exists a relabeling $\sigma\in S_n$ such that $A_{\sigma(1)}\cup A_{\sigma(2)}\cup\cdots\cup A_{\sigma(i)}$ has intersection with $A_{\sigma(i+1)}$. We can select $\sigma$ inductively. Pick $\sigma(1)$ arbitrarily. For simplicity, let $B_i = \cup_{j=1}^{i} A_{\sigma(j)}$. For any $i=1,\ldots,n-1$, if $B_i=[n]$ then pick $\sigma(i+1)$ arbitrarily; otherwise, take $S=B_i$ and by the lemma there exists $j$ such that $S$ has intersection with but does not contain $A_j$. Note that it is impossible to have $j=\sigma(1),\ldots,\sigma(i)$, so we can pick $\sigma(i+1)=j$. Therefore \begin{align*} A_1+\cdots+A_k&=B_{1}+A_{\sigma(2)}\cdots+A_{\sigma(k)}\\ &\supseteq (B_{1}\cap A_{\sigma(2)})+B_2+A_{\sigma(3)}+\cdots+A_{\sigma(k)}\\ &\supseteq (B_{1}\cap A_{\sigma(2)})+(B_2\cap A_{\sigma(3)})+B_3+\cdots+A_{\sigma(k)}\\ &\supseteq \ldots\\ &\supseteq (B_{1}\cap A_{\sigma(2)})+\cdots+(B_{n-1}\cap A_{\sigma(k)})+B_k. \end{align*}Since $B_{i}\cap A_{\sigma(i+1)}$ is non-empty for any $i=1,\ldots,k-1$ and $B_k=[n]$, we're done.