$1000$ does not satisfy the requirement, so let us consider only positive integers $n \le 999$. One can represent such an integer in the form $100a + 10b + c$ where $a, b, c$ are 1-digit non-negative integers. Since the number obtained by multiplying a 1-digit integer by $5$ is at most $45$ and its one's digit is either $0$ or $5$, we see that there is no carry-over in the addition $5a \cdot 100 + 5b \cdot 10 + 5c$. Hence if $n = a \cdot 100 + b \cdot 10 + c$, then $S(5n) = S(5a) + S(5b) + S(5c)$. Furthermore, if $k$ is an even 1-digit number, then $S(5k) = \frac{k}{2}$, while if $k$ is an odd 1-digit number, then $S(5k) = \frac{k+9}{2}$. Therefore, if $S(N) = S$ for a positive integer $N$ less than or equal to $999$, then $S(5N) = \frac{S+9\ell}{2}$, where $\ell$ is the number of odd integers among the one's, ten's and hundred's digits of $N$.

The remainder one gets when one divides a positive integer $n$ by $9$ equals the remainder one gets when $S(n)$ is divided by $9$. Hence if $S(N) = S(5N)$, then $N$ must be a multiple of $9$. There are $111$ multiples of $9$ less than or equal to $999$. We will find, by grouping them into sub-classes and checking, how many numbers among these will not satisfy the condition $S(5N) = S(N)$.

(1) When $S(N) = 9$.
Since $S(5N) = \frac{S(N)+9\ell}{2}$, where $\ell$ is the number of integers among the one's, ten's and hundred's digits of $N$, $\ell$ must be $1$ if $N$ in this class satisfies $S(5N) = S(N)$. Consequently, among those $N$'s in this class which do not satisfy the requirement have a representation corresponding with $(a, b, c)$ given by $(1, 1, 7), (1, 3, 5), (3, 3, 3)$ and their permutations, and there are exactly $3+6+1=10$ such numbers.

(2) When $S(N) = 18$.
Among $N$'s in this class those which satisfy the requirement have exactly $2$ odd integers among their one's, ten's and hundred's digits. If one such $N$ is written as $a \cdot 100 + b \cdot 10 + c$, then the number $999 - N = (9 - a) \cdot 100 + (9 - b) \cdot 10 + (9 - c)$ belongs to the class (1) and satisfies the requirement, and conversely, if $N' = a' \cdot 100 + b' \cdot 10 + c'$ is a number belonging to the class (1), which satisfies the requirement, then $999 - N' = (9 - a') \cdot 100 + (9 - b') \cdot 10 + (9 - c')$ belongs to class (2) and satisfies the requirement. Thus, there is a one-to-one correspondence between elements of class (1) and those of class (2), and the correspondence preserves the validity of the requirement. Therefore, as in class (1) there are $10$ numbers in class (2) which do not satisfy the requirement.

There is only the number $999$ belonging to this class, and this number satisfies the requirement.

Thus, among the multiples of $9$, there are $10 \times 2 = 20$ numbers which do not satisfy the requirement, and hence there are exactly $111 - 20 = 91$ positive integers $N$ less than or equal to $1000$ for which $S(5N) = S(N)$.