Let $n$ be a positive integer. There is a collection of cards that meets the following properties:
$\bullet$Each card has a number written in the form $m!$, where $m$ is a positive integer.
$\bullet$For every positive integer $t\le n!$, it is possible to choose one or more cards from the collection in such a way 
$\text{   }$that the sum of the numbers of those cards is $t$.
Determine, based on $n$, the smallest number of cards that this collection can have.
1. **Claim**: The smallest number of cards required is \(\frac{n(n-1)}{2} + 1\).

2. **Construction**: Consider the set of cards containing \(k\) copies of \(k!\) for \(k \in [1, n-1]\) and one card with \(n!\). This set satisfies the condition that for every positive integer \(t \leq n!\), it is possible to choose one or more cards such that the sum of the numbers on those cards is \(t\).

3. **Verification**: We need to show that this set of cards can indeed form any number \(t \leq n!\). We can use a greedy algorithm to achieve this:
   - Start with \(t\).
   - At each step, choose the largest factorial \(k!\) such that \(k! \leq t\).
   - Subtract \(k!\) from \(t\) and repeat until \(t\) becomes zero.

4. **Example**: Let's verify this for \(n = 3\):
   - The cards are \(1!\), \(2!\), and \(3!\).
   - To form \(5\), we can use \(3! = 6\) and \(1! = 1\), but since \(6 > 5\), we use \(2! = 2\) and \(1! = 1\) twice: \(2! + 1! + 1! = 2 + 1 + 1 = 4\), and then add another \(1!\) to get \(5\).

5. **Induction**: To show that at least \(\frac{n(n-1)}{2}\) cards are needed, we can use induction:
   - Base case: For \(n = 1\), we need \(1\) card, which is \(1!\).
   - Inductive step: Assume for \(n = k\), we need \(\frac{k(k-1)}{2} + 1\) cards. For \(n = k+1\), we need to add \(k+1\) copies of \((k+1)!\) to the existing set. This gives us \(\frac{k(k-1)}{2} + 1 + k + 1 = \frac{(k+1)k}{2} + 1\) cards.

6. **Conclusion**: The smallest number of cards required is \(\frac{n(n-1)}{2} + 1\).

The final answer is \( \boxed{ \frac{n(n-1)}{2} + 1 } \).