Let $T$ be the set of all 2-digit numbers whose digits are in $\{1,2,3,4,5,6\}$ and the tens digit is strictly smaller than the units digit. Suppose $S$ is a subset of $T$ such that it contains all six digits and no three numbers in $S$ use all six digits. If the cardinality of $S$ is $n$, find all possible values of $n$.
1. **Determine the size of set \( T \):**
   - \( T \) is the set of all 2-digit numbers whose digits are in \(\{1,2,3,4,5,6\}\) and the tens digit is strictly smaller than the units digit.
   - To form such a number, we choose 2 different digits from \(\{1,2,3,4,5,6\}\) and arrange them in increasing order.
   - The number of ways to choose 2 digits from 6 is given by \(\binom{6}{2}\).
   - Therefore, \(|T| = \binom{6}{2} = 15\).

2. **Check the feasibility for \( n < 4 \):**
   - If \( n < 4 \), it is impossible to include all six digits in \( S \) because each number in \( S \) uses exactly 2 digits.
   - Therefore, \( n \geq 4 \).

3. **Check the feasibility for \( n = 4 \) to \( n = 9 \):**
   - Consider \( S = \{12, 13, 14, 15, 16, 23, 24, 25, 26\} \).
   - This set \( S \) contains all six digits and no three numbers in \( S \) use all six digits.
   - By removing numbers from this set, we can create possible sets of every lower cardinality from 4 to 9.
   - Therefore, \( 4 \leq n \leq 9 \) are possible.

4. **Check the feasibility for \( n = 10 \):**
   - Suppose \( S \) has 10 numbers. We need to ensure that no three numbers in \( S \) use all six digits.
   - There are \(\frac{\binom{6}{2,2,2}}{3!} = 15\) "bad" sets of 3 numbers that contain all 6 digits.
   - Since there are 5 numbers not in \( S \) and each number is in exactly 3 different bad sets, we can eliminate \( 3 \times 5 = 15 \) bad sets at most.
   - This implies that we must eliminate exactly one number from each bad set.

5. **Case analysis for \( n = 10 \):**
   - Assume without loss of generality that \( 12 \in S \).
   - We need to eliminate one number from each of \(\{34, 56\}, \{35, 46\}, \{36, 45\}\).

   **Case 1: Eliminate 3 numbers with a shared digit:**
   - Suppose \( 34, 35, 36 \notin S \).
   - Remaining bad sets: \(\{13, 24, 56\}, \{13, 25, 46\}, \{13, 26, 45\}, \{14, 23, 56\}, \{15, 23, 46\}, \{16, 23, 45\}\).
   - The only numbers appearing 3 times in these bad sets are 13 and 23.
   - Eliminating these would result in \( S \) not containing the digit 3, which is a contradiction.

   **Case 2: Eliminate 3 numbers not sharing a digit:**
   - Suppose \( 34, 46, 36 \notin S \).
   - Remaining bad sets: \(\{13, 24, 56\}, \{13, 26, 45\}, \{14, 23, 56\}, \{14, 26, 35\}, \{16, 23, 45\}, \{16, 24, 35\}\).
   - There are no numbers appearing 3 times in these bad sets, so it is not possible to eliminate all of them with 2 numbers.
   - Therefore, \( n = 10 \) is not possible.

6. **Conclusion:**
   - Since every \( S \) with \( n > 10 \) has a subset with cardinality 10 that is not possible, there are no solutions for \( n \geq 10 \).
   - Therefore, the possible values of \( n \) are \( 4 \leq n \leq 9 \).

The final answer is \( \boxed{ 4 \leq n \leq 9 } \).