A set $X$ of positive integers is said to be [i]iberic[/i] if $X$ is a subset of $\{2, 3, \dots, 2018\}$, and whenever $m, n$ are both in $X$, $\gcd(m, n)$ is also in $X$. An iberic set is said to be [i]olympic[/i] if it is not properly contained in any other iberic set. Find all olympic iberic sets that contain the number $33$.
To solve the problem, we need to find all olympic iberic sets that contain the number \(33\). Let's break down the problem step by step.

1. **Definition of Iberic Set**:
   A set \(X\) of positive integers is iberic if \(X\) is a subset of \(\{2, 3, \dots, 2018\}\), and whenever \(m, n \in X\), \(\gcd(m, n) \in X\).

2. **Definition of Olympic Iberic Set**:
   An iberic set is olympic if it is not properly contained in any other iberic set.

3. **Initial Observations**:
   - The number \(33\) must be in the set \(X\).
   - The possible values of \(\gcd(33, x)\) for any \(x \in X\) are \(1, 3, 11, 33\).

4. **Exclusion of 1**:
   - If \(1\) were in \(X\), then \(X\) would not be a subset of \(\{2, 3, \dots, 2018\}\). Therefore, \(1 \notin X\).

5. **Inclusion of Multiples of 3 and 11**:
   - If \(\gcd(33, x) = 33\), then \(x\) must be a multiple of \(33\).
   - If \(\gcd(33, x) = 3\), then \(x\) must be a multiple of \(3\).
   - If \(\gcd(33, x) = 11\), then \(x\) must be a multiple of \(11\).

6. **Constructing Iberic Sets**:
   - Consider the set \(X_1 = \{3, 6, 9, 12, \dots, 2016\}\). This set contains all multiples of \(3\) up to \(2016\).
   - Consider the set \(X_2 = \{11, 22, 33, \dots, 2013\}\). This set contains all multiples of \(11\) up to \(2013\).

7. **Checking for Olympic Property**:
   - \(X_1\) is iberic because the \(\gcd\) of any two multiples of \(3\) is also a multiple of \(3\).
   - \(X_2\) is iberic because the \(\gcd\) of any two multiples of \(11\) is also a multiple of \(11\).
   - Both \(X_1\) and \(X_2\) are not properly contained in any larger iberic set that contains \(33\).

8. **Conclusion**:
   - The only olympic iberic sets that contain \(33\) are \(X_1\) and \(X_2\).

The final answer is \( \boxed{ \{3, 6, 9, 12, \dots, 2016\} } \) and \(\{11, 22, 33, \dots, 2013\}\).