Maximum number of 2 digit integers There are 90 cards with all two-digit numbers on them:
$$10,11,12,\dotsc,98,99$$
A player takes some of these cards. For each card taken she gets $1. However, if the player takes two cards that add up to 100 (say, 23 and 77), she loses all the money. How much could she get?
She can make at most $50.
Suppose there were 99 cards, numbered 1 to 99.  She would be able to take number 50, as well as one of 49 and 51, one of 48 and 52, ... and one of 1 and 99, for a total of 50 cards taken.  Because in this case she would be able to take cards the fifty cards numbered 50-99, she obviously can do so in our case as well (where cards 1-9 have been removed).  She can't make more money, because any additional card taken will pair with one of the cards she already took to total 100
If she has nine cards already, and all have partners on the table, then she has nine chances to lose 9 dollars and 72 chances to win 1, so she might as well stop.  The calculation changes depending on the number of unpaired cards she has.
If she has $m$ cards with partners on the table, and $n$ without partners, she has $m$ chances of losing $m+n$, and $90-2m-n$ chances of continuing, with $+1$ on that card.  The average is $\frac1{90}(-m^2-mn+90-2m-n)$, which is positive if $m+n\lt(90-m)/(m+1)$