Proof: Firstly we can simply assume $|A_i|=4$ If not,we can repalce $A_i$ by a $4-element$ subset of $A_i$ We randomly choose $24$ element from $X$ call it $Y=\{1,2,...24\}$ Of course,$Y$ may contain some $A_i$,denote $m$ by the number of $A_i$ such that $A_i$ is contained in $Y$,($m$ is obviously finite) All we need to do is to decrease $m$ WLOG let $A_1=\{1,2,3,4\}$ contained in $Y$ We delete $1$ can add an element from $\{25,26,...2009\}$ After deleting $1$,$m$ decrease by at least $1$,if we add $i$ into $Y$,$\{i,p,q,r\}$($i \in \{25,26,...2009\},p,q,r \in \{1,2,...24\}$) isn't any of $A_i$ Then we are done.So for any $i \in \{25,26,...2009\}$ $\exists p,q,r,A_j$ so that $A_j=\{i,p,q,r\}$ Let us count the possible $A_j$,It is easily to show there are at most $\binom{23}{3}$ such $A_j$(because $|A_i \cap A_j|\le 2$ and $|A_j|=4$) but from $\{25,26,...2009\}$ we are allowed to choose $1985$ which is bigger than $\binom{23}{3}$ Contradiction! QED

This is a new song on an old motif (triplets with at most one point in common). Take $| X | = n \geq 4$, and a maximal cardinality set $B$ with that property (such sets exist, e.g. any set of at most 3 elements); denote $| B | = b$. Then for any $x \in X \setminus B$ there exists at least one set $x \in A_i$ such that $A_i \setminus \{ x \} \subseteq B$ (otherwise we can append $x$ to $B$). Since $| A_i \setminus \{ x \} | \geq 3$, one may associate to $x$ any triplet $T_x \subseteq A_i \setminus \{ x \}$. For $y \neq x$, $y \in X \setminus B$, with set $A_j$ associated, one has $T_x \neq T_y$, or else $| A_i \cap A_j | \geq 3$. Therefore, by counting all possible triplets in $B$, one has $| X \setminus B | = n - b \leq \genfrac {(} {)} {0pt} {} {b} {3}$, leading to $b^3 - 3b^2 + 8b \geq 6n$. This is a strictly increasing polynomial in $b$ (its derivative is strictly positive), and one checks that $b \geq \lfloor \sqrt [3]{6n} \rfloor + 1$. For $n = 2009$ this yields $b \geq 24$.