Ascending digits

METHOD $1$ :

Positive Integers:

$0$ can not be used in any of these numbers as it will be put in the left most and that does not make any new number (it does not hold any value).
There are $9$ digits left from $1$ to $9$ . 1st we write them all in ascending order. Then we either take a digit or not take. So there are $2$ choices for each digit. So we get $2^{9}$ choices in total. But one of the choices is not taking any number which makes that number 0 which is not a positive integer (we skip it for now), so there are $2^{9} - 1$ positive integers.

Negative Integers:

Similarly there are $2^{9} - 1$ negative integers.

The only integer left is $0$ . So total integers $= 2^{9} - 1 + 2^{9} - 1 + 1 = 1023$

METHOD $2$ :

For positive integers,
total $1$ digit numbers, $^{9}$ C $_{1}$
total $2$ digit numbers, $^{9}$ C $_{2}$
$.....$
total $9$ digit numbers, $^{9}$ C $_{9}$
add all up to get $2^{9} - 1$
Similarly for negative integers we get $2^{9} - 1$
We also have to consider 0. So total $= 2^{9} - 1 + 2^{9} - 1 + 1 = 1023$

Gotcha I guessed it

Every non-empty subset of $\{1,2,3,4,5,6,7,8,9\}$ gives us exactly two "ascending integers". There are $2^{9}- 1 = 511$ such subsets, giving us $511$ positive and $511$ negative numbers. And there is $0$ . Thus there are $2 \times 511 + 1 = 1023$ such numbers altogether.