Generalizing my comment on Gareth's solution, we can
arrange Pascal's triangle as a right triangular array and ignore the right half ($n < 2k$) to obtain something like this:
1 4 6
We then, for any $N$,
Gareth has found the optimal solutions, but here is an R script if anyone wants to mess around with the upper bounds for n, just change the value of the variable"UpperBound".
gets to within about
of the desired answer. I think this is best possible with <= 100 cards.
Found with the help of a computer, but purely as an aid to calculation. My approach was to
[EDITED to add:]
Out of curiosity, I also ran a more automated search for the larger bound of n=500 mentioned in the OP. For this,
The automated search also ...