Enumerate the $k$-tuples; $(a_1, a_2, \ldots a_k)\rightarrow \binom{a_1}{1}+\binom{a_2-1}{2}+\binom{a_3-1}{3}+\ldots+\binom{a_k-1}{k}$ gives one method, where $a_1<a_2<a_3<\ldots<a_k$. Put $n$ into each of the $k$ sets with indices $a_i$ correlating to $n$. Then each $k$-tuple contains a unique element, and no number is contained in $k+1$ sets.

tenniskidperson3 wrote: Enumerate the $k$-tuples; $(a_1, a_2, \ldots a_k)\rightarrow \binom{a_1}{1}+\binom{a_2-1}{2}+\binom{a_3-1}{3}+\ldots+\binom{a_k-1}{k}$ gives one method, where $a_1<a_2<a_3<\ldots<a_k$. Put $n$ into each of the $k$ sets with indices $a_i$ correlating to $n$. Then each $k$-tuple contains a unique element, and no number is contained in $k+1$ sets. I really don't understand the above construction but it is not possible that a set in the family $ \mathcal{A} $ consists of finite number of elements. A construction of family $ \mathcal{A} $: When $k=2$ consider infinite many lines in a plane in general positions. When $k=3$ - infinite many planes in space in general positions will do the job. Of course a line(plane) consists of points with real coordinates, not in $\mathbb{N}$, but it just gives as a motivation. Let $\{a_i\}_{i=1}^{\infty}$ be be a sequence of real numbers which are pairwise different and $a_i \neq 0$. Let denote: $P_j = \left\{ (x_1,x_2,\ldots,x_{k}) \mid x_i\in \mathbb{R}, \sum_{i=1}^{k} a_j^i x_i -1 =0 \right\} $ $j=1,2,\ldots$. Now lets see that every $k$ different hyperplanes $P_{j_{\ell}},\, \ell=1,2,\ldots,k$ have exactly one common point $(x_1,x_2,\ldots,x_{k})$. This common point will satisfy the system $ \sum_{i=1}^{k} a_{j_{\ell}}^i x_i = 1 \,,\, \ell=1,2,\ldots, k $ But the determinant of the system is exactly the Vandermonde determinant $V=\prod_{t=1}^{k} a_{j_{t}}\prod_{1 \leq s < t \leq k} ( a_{j_{t}}-a_{j_{s}} ) \neq 0 $. So the above system has an unique solution. To see that every $k+1$ different planes have void intersection, assume on the contrary the hyperplanes $P_{j_{\ell}},\, \ell=1,2,\ldots,k+1$ have a common point $(x_1,x_2,\ldots,x_{k})$ Then the polynomial: $P(x)= x_1 x^1+x_2x^2+\ldots+x_k x^k-1$ will have $k+1$ different roots $a_{j_{\ell}},\, \ell=1,2,\ldots, k+1$, which means $P(x)$ is identically zero which is impossible. Now, it could have finished the proof but our sets must consist of natural numbers not of real $k$-tuples. But we can enumerate all possible intersections of $k$ different hyperplanes $P_j$ and let they be the points $\{p_{\ell}\}_{\ell=1}^{\infty}$. Now consider: $P'_j = \left\{\ell \mid \ell\in \mathbb{N},\, p_{\ell} \in P_j \right\}$ . These modified discrete sets will also satisfy the requirements (a) and (b).

My construction: For a number $n=p_1^{a_1}\cdot p_2^{a_2}\cdot ...\cdot p_m^{a_m}$ (its decomposition in prime factors) take $f(n)=a_1+a_2+...+a_m$. Now, for each prime $p$ take $A_p=\{px|\ x\in \Bbb{N},\ f(x)=k-1\}$. Now it's trivial to check the given conditions.

dgrozev wrote: I really don't understand the above construction but it is not possible that a set in the family $ \mathcal{A} $ consists of finite number of elements. Is this an issue? The problem statement didn't seem to require finite sets...

What I meant was that it's impossible some set in $\mathcal{A}$ to be finite. It can be easily shown. But, I don't know why I have decided that tenniskidperson3's construction produces finite sets. It was a long time ago. Sorry, apparently it was my fault. Maybe somehow I misunderstood the word "indices". Anyway, I would have put it in this way: Let $\mathcal{B}$ be the family of all finite subsets of $\mathbb{N}$ with exactly $k$ elements. Since $\mathcal{B}$ is countable, we can construct a bijection $f: \mathcal{B} \to \mathbb{N}$. Now, let us denote $A_k=\{j\in \mathbb{N}\mid k\in f^{-1}(j)\}\,, k\in \mathbb{N}$ and $\mathcal{A}=\{A_k \mid k\in \mathbb{N}\}$. Apparently $\mathcal{A}$ satisfies the problem's requirement.

Isn't this trivial... Just let $X$ be the set of square-free positive integers that have $k$ distinct prime divisors and let $A_i$ be the subset of $X$ that has the multiples of $p_i$ (the $i$-th prime). If we intersect $k$ sets, say $A_{a_1}$, ..., $A_{a_k}$ the intersection is $\{ p_{a_1} \times ... \times p_{a_k} \}$ and if we intersect $k+1$ sets the intersection is clearly empty since we would need $k+1$ distinct prime divisors.