Ninety-niner

Since $N$ is a multiple of 9, $a + b + c + d + e = 9p$ is a multiple of 9. It is easy to see that $1 \leq p \leq 5$ .

Since $N$ is a multiple of 11, $a - b + c - d + e = 11q$ is a multiple of 11. It is easy to see that $-1 \leq q \leq 2$ .

Adding/subtracting and dividing by two, we obtain $\begin{cases} a + c + e = \tfrac12(9p + 11q) \\ b + d = \tfrac12(9p - 11q)\end{cases}$ This requires that $p$ and $q$ have the same parity. Since digits lie between 1 and 9, we have the inequalities $\begin{cases} 6 - 11q \leq 9p \leq 54 - 11q \\ 4 + 11q \leq 9p \leq 36 + 11q \end{cases}$ Split out the possible values of $q$ and consider how much $p$ might be.

• $q = -1$ and $p$ is odd. The inequalities become $\begin{cases} 17 \leq 9p \leq 65 \\ -7 \leq 9p \leq 25 \end{cases}\ \ \ \therefore\ \ \ 2 \leq p \leq 2,$ which has no odd solutions.

• $q = 0$ and $p$ is even. $\begin{cases} 6 \leq 9p \leq 54 \\ 4 \leq 9p \leq 36 \end{cases}\ \ \ \therefore\ \ \ 1 \leq p \leq 4 \ \ \ \therefore\ \ \ p = 2, 4.$ Thus we have two solutions: $s(N) = 9\ \ \text{e.g.}\ N = 10\,098;\ \ \ \ \ \ s(N) = 18\ \ \text{e.g.}\ N = 19\,899.$

• $q = +1$ and $p$ is odd. $\begin{cases} -5 \leq 9p \leq 43 \\ 15 \leq 9p \leq 47 \end{cases}\ \ \ \therefore\ \ \ 2 \leq p \leq 4\ \ \ \therefore\ \ \ p = 3.$ This produces the solution $s(N) = \tfrac12(27 + 11) = 19\ \ \text{e.g.}\ N = 10\,989.$

• $q = +2$ and $p$ is even. $\begin{cases} -14 \leq 9p \leq 32 \\ 26 \leq 9p \leq 58 \end{cases}\ \ \ \therefore\ \ \ 3 \leq p \leq 3,$ which has no even solutions.

The possible values of $s(N)$ are therefore $9, 18, 19$ ; there are $\boxed{3}$ possible values.

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The last part of the solution can be illustrated in a graph as follows:

The orange and blue lines correspond to odd and even values, respectively, of $p$ and $q$ . Possible solutions lie on the crossing of lines of the same color. The requirements that $3 \leq a + c + e \leq 27$ and $2 \leq b + d\leq 18$ limit the solutions to the parallelogram defined by the black lines. The three solutions are marked by dots.

Mathematica

Union[Tr@IntegerDigits[#][[{1,3,5}]]&/@Table[99n,{n,102,1010}]]

{9, 18, 19}