Let $n$ be a positive integer. We want to make up a collection of cards with the following properties:

1. each card has a number of the form $m!$ written on it, where $m$ is a positive integer.

2. for any positive integer  $ t \le n!$, we can select some card(s) from this collection such that the sum of the number(s) on the selected card(s) is $t$.

Determine the smallest possible number of cards needed in this collection.
To determine the smallest possible number of cards needed in the collection, we need to ensure that any positive integer \( t \le n! \) can be represented as a sum of factorials of positive integers. This is closely related to the concept of factorial number system (factorial base).

1. **Understanding the Factorial Base Representation:**
   In the factorial number system, any positive integer \( t \) can be uniquely represented as:
   \[
   t = a_1 \cdot 1! + a_2 \cdot 2! + a_3 \cdot 3! + \cdots + a_n \cdot n!
   \]
   where \( 0 \le a_i \le i \) for \( i = 1, 2, \ldots, n \).

2. **Ensuring Coverage for All \( t \le n! \):**
   To cover all integers \( t \le n! \), we need to have enough cards such that we can form any sum up to \( n! \). This means we need:
   - 1 card of \( 1! \)
   - 2 cards of \( 2! \)
   - 3 cards of \( 3! \)
   - ...
   - \( n \) cards of \( n! \)

3. **Summing the Number of Cards:**
   The total number of cards required is the sum of the first \( n \) positive integers:
   \[
   1 + 2 + 3 + \cdots + n = \frac{n(n+1)}{2}
   \]

4. **Adding One Extra Card:**
   The problem statement suggests that we need one extra card to ensure that we can form any \( t \le n! \). This extra card is necessary to handle the case where the sum of the factorials exactly equals \( n! \).

Therefore, the smallest possible number of cards needed is:
\[
\frac{n(n+1)}{2} + 1
\]

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