This problem can be reduced to solving
$$x_1 + x_2 + x_3 = 15, \quad 0 \leq x_i \leq 9, \quad x_1 \neq 0, \quad x_1 \neq 9$$
First, we solve
$$x_1 + x_2 + x_3 = 15, \quad 0 \leq x_i \leq 15$$
This is a standard stars and bars problem, with solution $\binom{15+3-1}{15} = 136$. Now which solutions are wrong? Well, first of all those with one of them above $10$. But if, say, $x_1 \geq 10$, then we can write $x_1' = x_1 - 10$ and $x_2' = x_2, x_3' = x_3$ to reduce the problem to
$$x_1' + x_2' + x_3' = 5, \quad 0 \leq x_i' \leq 5$$
This is again a standard stars and bars problem, with solution $\binom{5+3-1}{5} = 21$. Since exactly one of $x_1,x_2,x_3$ could be above $10$, this gives $3 \cdot 21 = 63$ wrong solutions in total. So there are $136 - 63 = 73$ solutions to
$$x_1 + x_2 + x_3 = 15, \quad 0 \leq x_i \leq 9$$
Finally, we need to subtract $4$ solutions with $x_1 = 0$ (i.e. $(x_2,x_3) \in {(6,9),(7,8),(8,7),(9,6)}$), but also subtract $7$ solutions for $x_1 = 9$ (i.e. $(x_2,x_3) \in {(0,6),\ldots,(6,0)}$). This gives $62$ solutions in total.
Hint: How many different sum of digits are there for the cards?
(the smallest sum of digits is 1 from the card labeled 100, the largest is 27 from the card labeled 999).
Apply pigeon-hole principle to this result, noting that there is only one card with sum equal to 1 and only one card with sum equal to 27.
As there are 27 different possible sums on each card, pigeon-hole principle says that if you have $(27\cdot 2 + 1)$ cards, you are guaranteed to have three in the same category. This was based on the idea that you could have $(27\cdot 2)$ cards, with two in each category and avoiding having three in any single category. Note however, that you cannot have two in the category (sum=1) and you cannot have two in the category (sum=27). So, a final answer for this problem would be $(27\cdot2+1-2)=53$.
Best Answer
If the digit $7$ does not appear in a three-digit number, then the hundreds digit cannot be either $0$ or $7$, so there are eight choices for the hundreds digit. The only restriction on the tens digit is that it is not a $7$, so there are nine choices for the tens digit. Similarly, since the only restriction on the units digit is that is not a $7$, there are also nine choices for the units digit. Thus, there are $8 \cdot 9 \cdot 9$ three-digit positive integers which do not contain the digit $7$. Consequently, there are $$9 \cdot 10 \cdot 10 - 8 \cdot 9 \cdot 9 = 252$$ three-digit positive integers which contain at least one $7$ among its digits.