I'm not doing Newton's sum all the way!

Nope, I insist on doing Newton's sums all the way.

Let:

$S_1 = a_1 + a_2 + \ldots + a_9\\ S_2 = a_1a_2 + a_1a_3 + \ldots + a_8a_9\\ S_3 = a_1a_2a_3 + a_1a_2a_4 + \ldots + a_7a_8a_9\\ S_4 = a_1a_2a_3a_4 + a_1a_2a_3a_5 + \ldots + a_6a_7a_8a_9\\ S_5 = a_1a_2a_3a_4a_5 + a_1a_2a_3a_4a_6 + \ldots + a_5a_6a_7a_8a_9\\ S_6 = a_1a_2a_3a_4a_5a_6 + a_1a_2a_3a_4a_5a_7 + \ldots + a_4a_5a_6a_7a_8a_9\\ S_7 = a_1a_2a_3a_4a_5a_6a_7 + a_1a_2a_3a_4a_5a_6a_8 + \ldots + a_3a_4a_5a_6a_7a_8a_9\\ S_8 = a_1a_2a_3a_4a_5a_6a_7a_8 + a_1a_2a_3a_4a_5a_6a_7a_9 + \ldots + a_2a_3a_4a_5a_6a_7a_8a_9\\ S_9 = a_1a_2a_3a_4a_5a_6a_7a_8a_9$

Also, let

$P_1 = a_1 + a_2 + \ldots + a_9\\ P_2 = a_1^2 + a_2^2 + \ldots + a_9^2\\ P_3 = a_1^3 + a_2^3 + \ldots + a_9^3\\ \vdots\\ P_9 = a_1^9 + a_2^9 + \ldots + a_9^9\\ P_{10} = a_1^{10} + a_2^{10} + \ldots + a_9^{10}$

We know that:

$P_1 = S_1\\ P_2 = S_1P_1 - 2S_2\\ P_3 = S_1P_2 - S_2P_1 + 3S_3\\ P_4 = S_1P_3 - S_2P_2 + S_3P_1 - 4S_4\\ P_5 =S_1P_4 - S_2P_3 + S_3P_2 - S_4P_1 + 5S_5\\ P_6 =S_1P_5 - S_2P_4 + S_3P_3 - S_4P_2 + S_5P_1 - 6S_6\\ P_7 =S_1P_6 - S_2P_5 + S_3P_4 - S_4P_3 + S_5P_2 -S_6P_1 + 7S_7\\ P_8 = S_1P_7 - S_2P_6 + S_3P_5 - S_4P_4 + S_5P_3 -S_6P_2 + S_7P_1 - 8S_8\\ P_9 = S_1P_8 - S_2P_7 + S_3P_6 - S_4P_5 + S_5P_4 -S_6P_3 + S_7P_2 - S_8P_1 + 9S_9\\ P_{10} = S_1P_9 - S_2P_8 + S_3P_7 - S_4P_6 + S_5P_5 -S_6P_4 + S_7P_3 - S_8P_2 + S_9P_1$

Substitute the values one by one:

$\begin{cases}1 = S_1 & S_1=1\\ 2 = 1(1) - 2S_2 & S_2=-\dfrac{1}{2}\\ 3 = 1(2) +\dfrac{1}{2}(1) + 3S_3 & S_3=\dfrac{1}{6}\\ 4 = 1(3) +\dfrac{1}{2}(2) + \dfrac{1}{6}(1) - 4S_4 & S_4=\dfrac{1}{24}\\ 5 = 1(4) +\dfrac{1}{2}(3) + \dfrac{1}{6}(2) - \dfrac{1}{24}(1) + 5S_5 & S_5=-\dfrac{19}{120}\\ 6 = 1(5) +\dfrac{1}{2}(4) + \dfrac{1}{6}(3) - \dfrac{1}{24}(2) -\dfrac{19}{120}(1) - 6S_6 & S_6=\dfrac{151}{720}\\ 7 = 1(6) +\dfrac{1}{2}(5) + \dfrac{1}{6}(4) - \dfrac{1}{24}(3) -\dfrac{19}{120}(2) -\dfrac{151}{720}(1) + 7S_7 & S_7=-\dfrac{1091}{5040}\\ 8 = 1(7) +\dfrac{1}{2}(6) + \dfrac{1}{6}(5) - \dfrac{1}{24}(4) -\dfrac{19}{120}(3) -\dfrac{151}{720}(2) -\dfrac{1091}{5040}(1) - 8S_8 & S_8=\dfrac{7841}{40320}\\ 9 = 1(8) +\dfrac{1}{2}(7) + \dfrac{1}{6}(6) - \dfrac{1}{24}(5) -\dfrac{19}{120}(4) -\dfrac{151}{720}(3) -\dfrac{1091}{5040}(2) - \dfrac{7841}{40320}(1) + 9S_9 & S_9=-\dfrac{56519}{362880}\\ P_{10} = 1(9) +\dfrac{1}{2}(8) + \dfrac{1}{6}(7) - \dfrac{1}{24}(6) -\dfrac{19}{120}(5) -\dfrac{151}{720}(4) -\dfrac{1091}{5040}(3) - \dfrac{7841}{40320}(2) -\dfrac{56519}{362880}(1) & =\dfrac{4025071}{362880}\\ \end{cases}$

Note: I verified that this value is the correct one in its simplest form with the help of WolframAlpha

Now, we are looking for the smallest possible value of $k$ such that $kP_{10} = \dfrac{4025071k}{362880}$ is an integer

Obviously, the answer would be $\boxed{362880}$

The truth:

Most of us who solved this question didn't actually computed this value. (I probably was the only person who did this, I actually computed this after I solved the question, just to do Newton's sums all the way and because this is the only way I know how to prove my answer)

What I did was, I computed all the way to $S_4$ , and I then realized that $P_{10}$ probably was:

$P_{10} = P_1 + \dfrac{P_2}{2!} + \dfrac{P_3}{3!} - \dfrac{P_4}{4!} \pm \dfrac{aP_5}{5!} \pm \dfrac{bP_6}{6!} \pm \dfrac{cP_7}{7!} \pm \dfrac{dP_8}{8!} \pm \dfrac{eP_9}{9!}$

where $a,\;b,\;c,\;d,\;e$ are all constants.

I guessed that $P_{10} = \dfrac{n}{9!}$ , where $n$ and $9!$ are coprime positive integers. Therefore, for $kP_{10} = \dfrac{kn}{9!}$ to be an integer, the smallest positive value that $k$ can take is probably $9! = \boxed{362880}$ . Hopefully, someone else can write a solution which can prove this.