Let $X$ satisfy the conditions of the problem, and without loss of generality, let $A_{1}=\{1,2,3\}$. By condition (2), there exists an element of $X$ that does not contain 1, and we can assume $A_{2}=\left\{a_{1}, a_{2}, a_{3}\right\}$, and $1 \notin A_{2}$. Similarly, we can set $A_{3}=\left\{b_{1}, b_{2}, b_{3}\right\}$, and $2 \notin A_{3}$, $A_{4}=\left\{c_{1}, c_{2}, c_{3}\right\}$, and $3 \notin A_{4}$. By condition (1), each element of $X$ must contain at least one of the following 9 pairs as a subset:
$\left\{1, a_{1}\right\},\left\{1, a_{2}\right\},\left\{1, a_{3}\right\},\left\{2, b_{1}\right\},\left\{2, b_{2}\right\},\left\{2, b_{3}\right\},\left\{3, c_{1}\right\},\left\{3, c_{2}\right\},\left\{3, c_{3}\right\}$. Since $|I|=36$, each of these pairs can be a subset of at most $36-2=34$ elements of $X$.

If there do not exist 3 pairs such that each pair is a subset of at least 4 elements of $X$, then at most 2 pairs can be subsets of 34 elements of $X$, and the remaining 7 pairs can be subsets of at most 3 elements of $X$. Therefore,
$n=|X| \leqslant 2 \times 34 + 7 \times 3 = 89$.
Now assume that among these 9 pairs, there are at least 3 pairs such that each pair is a subset of at least 4 elements of $X$. Let these 3 pairs be $p_{1}=\{a, b\}, p_{2}=\{c, d\}, p_{3}=\{e, f\}$.
Since $p_{1}$ is a subset of at least 4 elements of $X$, $\{a, b, x\},\{a, b, y\},\{a, b, z\},\{a, b, w\}$, and any other element $A$ of $X$ must share at least one common element with these 4 elements, which can only be $a$ or $b$ (otherwise, $A$ would contain 4 different elements $x, y, z, w$, which is a contradiction). Similarly, this applies to $p_{2}$ and $p_{3}$.

If $p_{1} \cap p_{2}=\varnothing$, then there must be two elements of $X$ of the form $\{a, b, x\}$ and $\{c, d, y\}$ (where $x \neq c, d, y, y \neq a, b, x$), which contradicts condition (1). Therefore, $p_{1}, p_{2}, p_{3}$ must have non-empty intersections.
We consider two cases:
( I ) $p_{1} \cap p_{2} \cap p_{3} \neq \varnothing$, let $p_{1}=\{x, y\}, p_{2}=\{x, z\}, p_{3}=\{x, w\}$. Here, $x, y, z, w \in I$. Thus, the elements of $X$ that do not contain $x$ can only be $\{y, z, w\}$. Therefore, $\{y, z, w\}$ is an element of $X$ and $X \subseteq Y(x, y, z, w)$, where $Y(x, y, z, w)$ is the set of all 3-element subsets that contain $x$ and at least one of $y, z, w$. Thus, $n=|X| \leqslant |Y(x, y, z, w)| = 1 + C_{3}^{2} + C_{3}^{1} C_{32}^{1} = 100$, with equality holding if and only if $X = Y(x, y, z, w)$. There are $C_{36}^{3} \cdot C_{33}^{1} = C_{36}^{1} \cdot C_{35}^{3} = 33 \cdot C_{36}^{3}$ such $X = Y(x, y, z, w)$.
( II ) $p_{1} \cap p_{2} \cap p_{3} = \varnothing$. Since $p_{1}, p_{2}, p_{3}$ have non-empty intersections, we can set $p_{1}=\{x, y\}, p_{2}=\{y, z\}, p_{3}=\{x, z\}$, where $x, y, z \in I$. Thus, each element of $X$ must contain at least 2 of $x, y, z$, so $X \subseteq Y(x, y, z)$, where $Y(x, y, z)$ is the set of all 3-element subsets that contain at least 2 of $x, y, z$. Thus, $n=|X| \leqslant |Y(x, y, z)| = 1 + C_{3}^{2} C_{33}^{1} = 100$, with equality holding if and only if $X = Y(x, y, z)$. There are $C_{36}^{3}$ such $X = Y(x, y, z)$, and they are different from those in case (I).

In summary, the maximum value of $n$ is 100, and when $n$ reaches its maximum value, there are $33 C_{36}^{3} + C_{36}^{3} = 34 C_{36}^{3}$ different $X$.
Example 14 Let $X=\left\{A_{1}, A_{2}, \cdots, A_{n}\right\}$ be a family of distinct three-element subsets of $I=\{1,2,3, \cdots, 36\}$, satisfying:
(1) For any $1 \leqslant i<j \leqslant n, A_{i} \cap A_{j} \neq \varnothing$,
(2) $A_{1} \cap A_{2} \cap \cdots \cap A_{n}=\varnothing$.

Find the maximum value of $n$, and how many different families $X$ are there when $n$ reaches its maximum value?