Selection minimum out of $n$ different objects!

Question

Suppose we have $n$ differnt persons and $n$ different objects.They have to select from the the objects such that every pair of person has at least one uncommon object.
What is the minimum number of objects that they should select?
EDIT
Its also given that every person can select as many objects as he wants.
But every pair should have at least one uncommon object each.
So find the total minimum number of objects that they should select out of n. CLARIFICATION
For ever pair of person $A$ and $B$, $A$ must have an object that $B$ doesn't have and $B$ must have an object that $A$ doesnt have.
EXAMPLE
For $n=5$ there are $5$ persons $\{A,B,C,D,E\}$ and objects numbered $\{1,2,3,4,5\}$.
The min. number of objects they should select is $4$.
$$\begin{align*} A: & (3,1)\\ B: & (3,2)\\ C: & (3,4)\\ D: & (4,1)\\ E: & (1,2) \end{align*}$$ This is one possible combination, but the minimum total is $4$ elements.

Answer 1

Reformulating the question to match the example (OP: copy this if it's correct)

For a number of $n$ persons, who pick a number of Objects $A_i \subset \Omega$.
What is the minimum number of elements of $\Omega$, such that the persons can pick objects fulfilling the following condition? $$A_i \setminus A_j \neq \emptyset \qquad \forall\ i \neq j \in \{1, \ldots, n\}$$

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Ansatz based on Michaels' observations

Let $N$ be even, so we don't have to worry about $\lfloor \cdot \rfloor$. Then we want $$\begin{align*}n \lesssim \binom{N}{N/2} & = \frac{N(N-1)(N-2) \ldots (N/2)}{(N/2)(N/2 - 1) \ldots} \\ & = \prod_{k=0}^{N/2-1} \frac{N-k}{N/2 - k} = 2^{N/2} \prod_{k=1}^{N/2-1} \frac{N-k}{N-2k} \\ & = 2^{N/2} \prod_{k=1}^{N/2 - 1} \left(1 + \frac{k}{N-2k}\right) \end{align*}$$ Where $\lesssim$ denotes, that $N$ should be chosen such, that $N-1$ doesn't fulfill the $<$ restriction, but $N$ does.

Answer 2

In OP's example, there could be person $F$ with objects 2 and 4. So there are ${4\choose 2}$ people choosing two from four objects.

Suppose there are $N$ objects, and everyone chooses $k$ objects.

To answer the opposite question from the OP, what is the maximum number of people that can choose from $N$ objects, it would be ${N\choose\lfloor N/2 \rfloor}$ people.