40. Let the set $X$ have $n>1$ elements, and there is a relation $\wedge$, satisfying:
(1) For any $x \in X, x \wedge x$ does not hold;
(2) For any pair of distinct elements $x, y \in X, x \wedge y$ and $y \wedge x$ hold exactly one of them;
(3) If $x \wedge y$, then there exists $z \in X$ such that $x \wedge z, z \wedge y$,
How many elements does $X$ have at least?
40. Let $x_{1}, x_{2} \in X$. By (2), we can assume $x_{1} \wedge x_{2}$. By (3), there exists $x_{3} \in X$ such that $x_{1} \wedge x_{3} \wedge x_{2}$ (i.e., $x_{1} \wedge x_{3}, x_{3} \wedge x_{2}$).
Similarly, there exist $x_{4}, x_{5}, x_{6}, x_{1} \in X$ satisfying:
$x_{1} \wedge x_{4} \wedge x_{3}, x_{4} \wedge x_{3} \wedge x_{3}, x_{3} \wedge x_{1} \wedge x_{2}, x_{3} \wedge x_{5} \wedge x_{6}$.
By (1), $x_{3} \neq x_{1}, x_{2}$, and $x_{1} \neq x_{1}, x_{3}$. By (2), $x_{1} \neq x_{2}$.
Similarly, $x_{5} \neq x_{1}, x_{2}, x_{1}, x_{1}$; $x_{6} \neq x_{1}, x_{2}, x_{3}, x_{1}, x_{5}$; $x_{i} \neq x_{1}, x_{2}, x_{3}, x_{4}, x_{5}, x_{1}$.
Thus, $X$ has at least 7 elements.
On the other hand, for any finite set $X$ with at least 7 elements, a binary relation $\wedge$ can be established to satisfy (1), (2), and (3).
Case 1: $X=\{1,2, \cdots, n\}$, where $n$ is an odd number $(\geqslant 7)$.
For $1 \leqslant s \leqslant n-4$, $x \wedge (x-n+4) \wedge y$. 
$2^{\circ}$ If $y-x$ is a positive even number less than $n-1$, when $y-x>2$, $x \wedge (x \div 2) \wedge y$; when $y-x=2$, $x \wedge (x+1) \wedge y$.
$3^{\circ}$ If $x-y=n-1$ or a positive odd number greater than 1, when $y \geqslant 3$, $x-y$ is a positive odd number greater than 1, $x \wedge (y-2) \wedge y$; when $x \leqslant n-2$, $x \wedge (x+2) \wedge y$; when $y=1$ and $x=n-1$, $x \wedge n \wedge y$; when $y=1$ and $x=n$, $x \wedge (n-3) \wedge y$; when $y=2$ and $x=n$, $x \wedge 1 \wedge y$.
Thus, (3) holds.
Case 2: $X=\{1,2, \cdots, n+1\}$, where $n$ is an odd number $(\geqslant 7)$.
Define $\wedge$ on $\{1,2, \cdots, n\}$ as in Case 1. Additionally, $x \wedge (n+1)$ for $x=1, 2, \cdots, n$.

Clearly, (1) and (2) hold. Suppose $x \wedge y$. If $x, y \leqslant n$, (3) holds as in Case 1. If $y=n+1$ and $x \leqslant n-1$, then $x \wedge (x+1) \wedge y$. If $y=n+1$ and $x=n$, then $x \wedge 1 \wedge y$. Therefore, (3) holds.
In conclusion, the minimum number of elements in $X$ is 7.