3. Consider the set $S$ of all integer points (points with integer coordinates) on the coordinate plane. For each positive integer $k$, if there exists a point $C \in S$ such that the area of $\triangle A B C$ is $k$, then two distinct points $A, B \in S$ are called “$k$-friends”; if any two points in $T$ are $k$-friends, then the set $T \subset S$ is called a “$k$-clique”. Find the smallest positive integer $k$ such that there exists a $k$-clique with more than 200 elements.
3. First, consider the point $B \in S$ that is a $k$-friend of the point $(0,0)$. Let $B(u, v)$. Then $B$ is a $k$-friend of $(0,0)$ if and only if there exists a point $C(x, y) \in S$ such that $\frac{1}{2}|u y - v x| = k$ (here we use the area formula of $\triangle A B C$, where $A$ is the origin).

There exist integers $x, y$ such that $|u y - v x| = 2k$ if and only if the greatest common divisor (gcd) of $u, v$ is a divisor of $2k$. Therefore, the point $B(u, v) \in S$ is a $k$-friend of $(0,0)$ if and only if the gcd of $u, v$ divides $2k$.

If two points are $k$-friends, translating them to other points in $S$ will still make them $k$-friends. Thus, two points $A, B \in S, A(s, t), B(u, v)$ are $k$-friends if and only if the point $(u-s, v-t)$ is a $k$-friend of $(0,0)$, i.e., the gcd of $u-s, v-t$ divides $2k$.

Let $n$ be a positive integer that does not divide $2k$. Then the number of elements in a $k$-clique is at most $n^2$.

In fact, all points $(x, y) \in S$ are divided into $n^2$ classes modulo $n$. If a set $T (T \subset S)$ contains more than $n^2$ elements, then there exist two points $A, B \in T$ that belong to the same class. Let $A(s, t), B(u, v)$. Then
$n \mid (u-s), n \mid (v-t)$.
Thus, $n \mid d$, where $d$ is the gcd of $u-s, v-t$. Since $n$ does not divide $2k$, $d$ does not divide $2k$, and $A, B$ are not $k$-friends, so $T$ is not a $k$-clique.

Let $M(k)$ be the smallest positive integer that does not divide $2k$. Denote $M(k) = m$.

Consider the points $(x, y) \in S$ satisfying $0 \leqslant x, y \leqslant 200$.

By the definition of $M(k)$, $2k$ can be divided by $1, 2, \cdots, M(k)-1$, but not by $M(k)$. If $(M(k))^2 > 200$, then $M(k) \geqslant 15$.
(1) $M(k) = 15$, which clearly contradicts the fact that $2k$ can be divided by 3 and 5, but not by 15.
(2) $M(k) = 16$, then $2k$ can be divided by $1, 2, \cdots, 15$, and also by their least common multiple (LCM) $L$, but not by 16. Since $L$ is not a multiple of 16, $k = \frac{L}{2}$ is the smallest $k$ such that $M(k) = 16$.
(3) $M(k) \geqslant 17$, then $2k$ can be divided by the LCM of $1, 2, \cdots, 16$, which is $2L$. From $2k \geqslant 2L$, we get $k \geqslant L > \frac{L}{2}$.
In summary, the smallest $k$ that satisfies the condition is
$$
\frac{L}{2} = 180180
$$