Decks are composed of 1 copy of each of $D$ unique cards. The set of cards is $C$ ($|C|=D$), the set of people is $P$ ($|P|=n\geq k$).


Starting with a simpler case (dropping the $k-1$ restriction)
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One answer is to give $n-k+1$ full decks to $n-k+1$ people. Then applying pigeonhole principle, any $k$ people will definitely have a full deck. This takes $n-k+1$ decks and $D\times (n-k+1)$ cards. $\blacksquare$


We can show this is minimum. Suppose there are $\leq n-k$ of some card $c\in C$ and hence $\geq n-(n-k)=k$ people who don't have $c$. So, those $k$ people cannot assemble a full deck. Hence, at least $n-k+1$ of each card is required and hence $n-k+1$ full decks. $\blacksquare$


Main Problem
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Even adding in the restriction that any $k-1$ people cannot together have a full deck, the previous minimum still holds. i.e. we still need $\geq n-k+1$ full decks.


Possible general solution.
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There are $\left(\frac{n}{k-1}\right)$ groups of $k-1$ people. This strategy cannot work with $D<\left(\frac{n}{k-1}\right)$.


Partition the set of cards $C$ into $\left(\frac{n}{k-1}\right)$ sets. For extra "niceness" we can easily ensure these parts will have size $\in \left \{ \left \lceil\dfrac{D}{\left(\frac{n}{k-1}\right)}\right \rceil, \left \lfloor\dfrac{D}{\left(\frac{n}{k-1}\right)}\right \rfloor \right \}$


For every group $G\_i$ of $k-1$ people, assign a unique part $C\_i$ to this group, and take away the cards from $C\_i$ from each member of $G\_i$.


For groups $G\_i$ and $G\_j$, $i\neq j$:
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By construction $G\_i \setminus G\_j \neq \emptyset \implies \exists g\in G\_i, g\notin G\_j$. So $g$ has all the cards of $C\_j$.


So members of $G\_i$ are together only missing the cards of $C\_i$.


By construction $\forall g\notin G\_i\implies g$ has cards of $C\_i$. $\blacksquare$


Number of cards used
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Number of cards is optimal if $D>\left(\frac{n}{k-1}\right)$ as it achieves the previously computed lower bound. $\blacksquare$


Questions
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Im assuming (perhaps prematurely) that my solution is correct.


**Can we prove that there is no solution for the case $D<\left(\frac{n}{k-1}\right)$?**



How to distribute least number of $D$ card decks amongst $n$ people so that any $k$ people have a full deck and no $k-1$ people have a full deck
For each $G\subseteq P$ define $C\_G\equiv\{$**Cards not held by any member of G**$\}$.


Notice $\forall G,H\subseteq P, C\_{G\cup H}=C\_G\cap C\_H$


Let $K\equiv\{G\subseteq P : |G|=k-1\}$. Notice $|K|=\left (\frac{n}{k-1}\right )$ [By definition of $\left (\frac{n}{r}\right )$]


Claim
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$\forall G\in K, C\_G\neq \emptyset$


Proof
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$|G|=k-1$. But $C\_G=\emptyset\implies G$ has a full deck. $\blacksquare$


Claim
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$\forall G,H\in K, G\neq H \implies C\_G \cap C\_H = \emptyset$.


Proof
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Let $\exists G,H\in K, G\neq H, C\_G \cap C\_H \neq \emptyset$


Consider $h\in H$.


By construction, $C\_{\{h\}}\supseteq C\_H \supseteq C\_G \cap C\_H \neq \emptyset$.


$\therefore C\_{G\cup\{h\}}=C\_G\cap C\_{\{h\}} \supseteq C\_G \cap C\_G \cap C\_{\{h\}}\neq \emptyset$.


$\implies C\_{G\cup\{h\}} \supset \emptyset \implies C\_{G\cup\{h\}} \neq \emptyset$


$|G\cup\{h\}|=k$. $G\cup\{h\}$ does not have a full deck.$\blacksquare$


Putting it together
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Maximum size partition of $C$ is $|C|=D$.


Notice the set $R\equiv\{C\_G : G\in K\}$ forms a partition of $D$. [See 2 above claims]


$|R|=|K| \implies |K|=\left (\frac{n}{k-1}\right )\leq D$ [Pigeonhole principle rears its head once more!] $\blacksquare$


