Those are some nifty numbers

With the exception of $9999 = 10000 - s(10000)$ , any positive integer $m$ which is a nifty number can be written as $n - s(n)$ where $n$ is at most a $4$ -digit number. If $n = \overline{abcd}$ is an at most $4$ -digit number, then $n-s(n) = 999a + 99b + 9c = 9(111a + 11b+c)$ . The value of $d$ does not matter, so we might as well choose $d=1$ , so that $n$ is always a positive integer, for any values of $0 \le a,b,c \le 9$ .

Now the map $(a,b,c) \mapsto 111a + 11b + c \hspace{2cm} a,b,c \,=\, 0,1,2, \ldots 9$ is injective, and hence there are $10^3 = 1000$ numbers of the form $999a + 99b + 9c = n-s(n)$ for a positive integer $n < 10000$ . We need to exclude the case $a=b=c=0$ , since this gives the nifty number $0$ , which is not positive. On the other hand, we need to include the nifty number $9999$ in our count. Thus there are $\boxed{1000}$ nifty positive integers less than $10000$ .

Let $f(n)=n-s(n)$ . Observe that

$f(10m)=f(10m+1) = f(10m+2) = f(10m+3)=\ldots =f(10m+9)$

for all positive integers $m$ . From this, we get all nifty numbers belong in the sequence $f(10), f(20), f(30), \ldots$ .

Also note $f(10m) < f(10m+10)$ for all positive integers $m$ . This statement is equivalent to

$10m - s(10m) < 10m+10 - s(10m+10) \iff s(m+1)-s(m) < 10$

where we used the fact $s(10m)=s(m)$ . The inequality sign $s(m+1)-s(m) < 10$ is trivially true for $m \not \equiv 9 \pmod{10}$ , since $m+1$ and $m$ only differ in their last digit. To see that $s(m+1) - s(m) < 10$ if $m \equiv 9 \pmod{10}$ , suppose $m$ ends in $k$ 9s. Then the last $k$ digits of $m+1$ are 0's, while the preceding digit has increased by 1. Thus, $s(m+1) - s(m) = 1-9k < 10$ .

Since $f(10000)=9999$ and $f(10010)=10008$ , it follows that the nifty positive integers less than 10000 are precisely the following.

$f(10) < f(20) < f(30) < \ldots < f(10000)$

Therefore, there are $\boxed{1000}$ nifty positive integers less than 10000.