Any three sets have empty intersection -- how many sets can there be?

Let $\mathcal F$ be a family of nonempty subsets of $X=\{1,2,\ldots, N\}$, such that for any three distinct $A,B,C \in \mathcal F$ we have $A \cap B \cap C = \varnothing$. What is the maximum cardinality of $\mathcal F$?

I suspect the answer is about $3N/2$ and can be obtained from for example $$\mathcal F = \{\{1\},\{2\},\ldots , \{N\}, \{1,2\}, \{3,4\},\ldots, \{N-1,N\}\} \text{ assuming even } N.$$

But so far I can only prove the upper bound of $2N-1$. This is obtained by considering the function that takes every set in $\mathcal F$ to its minimum element. The fibres over $1,2,\ldots, N-1$ can each contain at most two sets, as otherwise we would get three sets with a shared element, which is not allowed. The fibre over $N$ can contain at most the set $\{N\}$. Hence we get the bound $2(N-1)+1=2N-1$.

Yes, the maximum possible size of $\newcommand\F{\mathcal F}\F$ is $\lfloor 3n/2\rfloor$.

Consider the $n\times |\F|$ incidence matrix for this set family. Each row corresponds to a number in $\{1,\dots,n\}$, and each column corresponds to a subset in $\F$. The entry is $1$ if the subset contains that number, and $0$ otherwise.

Let us count the number of ones in this matrix in two ways. For each $k\in \{1,\dots,n\}$, let $f_k$ be the number of subsets in $\F$ with cardinality $k$. On the one hand, there are exactly $f_1+2f_2+3f_3+\dots+nf_n$ ones in the matrix, counting by columns. On the other hand, the condition that $A\cap B\cap C=0$ implies there are at most two ones in each row, so at most $2n$ ones total. This proves $$ f_1+2f_2+\dots+nf_n\le 2n $$ Conclude as follows: $$ \begin{align} |\F| &=f_1+f_2+\dots+f_n \\&= \frac12f_1+ \frac12(f_1+2f_2+2f_3+\dots+2f_n) \\&\le \frac12f_1+\frac12(f_1+2f_2+\color{blue}3f_3+\dots+\color{blue}nf_n) \\&\le \frac12\cdot n+\frac12\cdot 2n \end{align} $$ This inequality, combined with the fact that $|\F|$ is an integer, implies $|\F|\le \lfloor 3n/2\rfloor$. The construction in your post achieves this bound exactly, so it is optimal.

We note that we can pick $\mathcal{F}$ to consist only of sets of size one or two, for if $\mathcal{F}$ contains $A = \{a_1,\dots, a_r\}$ for $r\geq 3$, we can replace $A$ by either $\{a_1\}$ or $\{a_1, a_2\}$, as at least one of these is not in $\mathcal{F}$. This maintains the triple intersection property. We can also choose $\mathcal{F}$ to be downward closed, as replacing a set by one of its subsets not in $\mathcal{F}$ maintains the triple intersection property. Finally, in such an $\mathcal{F}$ any 2-element sets are disjoint. This shows your example is optimal.

Let $m$ be the maximum possible size of such a family. Among all such families of size $m$, consider a family $\mathcal F$ which minimizes the quantity $\sum_{X\in\mathcal F}|X|$.

Note that, if $X\in\mathcal F$, then all nonempty subsets of $X$ must also belong to $\mathcal F$; for if some nonempty subset $Y$ of $X$ did not belong to $\mathcal F$ we could replace $X$ with $Y$.

It follows that every element of $\mathcal F$ has at most two elements; for if a set $X\in\mathcal F$ contained three distinct elements $a.b.c$ then $\mathcal F$ would contain four sets $\{a,b,c\},\{a,b\},\{a,c\},\{a\}$ with nonempty intersection. Moreover, the $2$-element member of $\mathcal F$ must be pairwise disjoint; if $\{a,b\}\in\mathcal F$ and $\{a,c\}\in\mathcal F$, then $\mathcal F$ contains three sets $\{a,b\},\{a,c\},\{a\}$ with nonempty intersection.

Hence $\mathcal F$ contains at most $N$ $1$-element sets and at most $\left\lfloor\frac N2\right\rfloor$ $2$-element sets, so that $m=|\mathcal F|\le N+\left\lfloor\frac N2\right\rfloor=\left\lfloor\frac{3N}2\right\rfloor$.

Remark. Likewise, if the condition is weakened to "Any four sets have empty intersection," then (for $N\ge3$) the maximum cardinality of $\mathcal F$ is $2N$, which is attained by taking the sets $\{1\},\{2\},\dots,\{N\}$ and $\{1,2\},\{2,3\},\dots,\{N-1,N\},\{N,1\}$.

For $S \subseteq \{1,\dots,n\}$, let binary decision variable $x_S$ indicate whether $S \in \mathcal F$. The problem is to maximize $\sum_S x_S$ subject to linear constraints $$\sum_{S: i \in S} x_S \le 2 \quad \text{for all $i\in\{1,\dots,n\}$} \tag1\label1$$ that prevent each $i$ from appearing more than twice.

It turns out that the linear programming (LP) relaxation obtained by ignoring integrality of $x_S$ yields maximum objective value $3n/2$, which implies an upper bound of $\lfloor 3n/2 \rfloor$ for the original problem. The LP dual variables provide a short certificate of optimality. Explicitly, take dual variable $1/2$ for the upper bound constraint $x_S \le 1$ on each $S$ such that $|S|=1$, dual variable $|S|/2-1$ for the lower bound constraint $-x_S \le 0$ on each $S$ such that $|S|\ge 2$, and dual variable $1/2$ for each constraint in \eqref{1}, yielding \begin{align} \sum_S x_S &= \sum_S \left(1 - \frac{|S|}{2} + \frac{|S|}{2}\right) x_S \\ &= \sum_S \left(1-\frac{|S|}{2}\right) x_S + \frac{1}{2}\sum_S \sum_{i\in S} x_S \\ &= \sum_{S:|S|=1} \left(1-\frac{|S|}{2}\right) x_S + \sum_{S:|S|\ge 2} \left(1-\frac{|S|}{2}\right) x_S + \frac{1}{2}\sum_S \sum_{i\in S} x_S \\ &= \sum_{S: |S|=1} \frac{1}{2} x_S + \sum_{S: |S| \ge 2} \left(\frac{|S|}{2}-1 \right) (-x_S) + \sum_{i=1}^n \frac{1}{2} \sum_{S:i\in S} x_S \\ &\le \sum_{S: |S|=1} \frac{1}{2} \cdot 1 + \sum_{S: |S| \ge 2} \left(\frac{|S|}{2}-1\right) 0 + \sum_{i=1}^n \frac{1}{2} \cdot 2 \\ &= \frac{n}{2} + 0 + n \\ &= \frac{3n}{2}. \end{align}

In retrospect, this is essentially the same argument as in Mike Earnest's answer. The correspondence is $f_k = \sum_{S: |S|=k} x_S$. Note that the LP dual variables arise automatically as a by-product of an LP solver call.