Pack having unnamed cards............

Because the cards are ascending natural numbers, we know that the sum of all of them must be a triangular number. Let k be the lower of our integers. Therefore, 1224+k+(k+1) must equal a triangular number. The closes triangular number>1224 is 1225 (49 cards in the deck;49th triangular number). This obviously does not work, because then the two cards would have to sum to 1, and 0 is not a card in the set. The next highest is 1275 (50 cards). The difference between 1275 and 1224 is 51, which can be written as 26+25. 25 being the smaller, $\boxed { 25-20=5 }$

N.B: Essentially, we are trying to find values satisfying the equation $\frac { n(n+1) }{ 2 } =1224+(2k+1)$ . There are infinite solutions natural number solutions to this. However, you will notice that all other solutions do not fit with the problem. e.g.: $\frac { 53(54) }{ 2 } =1224+207$ . While this does satisfy the equation, it is problematic because you would have to remove the two consecutive cards 103 and 104. This is a contradiction, because the deck in this case would have only 53 cards.

For $n=50$ , we have

$\frac{n (n+1)}{2}=1275$

Taking $K=25$ ,

$K+K+1=51$ and $1275 - 51 = 1224$

$K-20=5$