..............................................................................................................................................

Indeed, $9999=10^4-1$ is the minimum value of $k$. First, suppose that $k$ have one digite. Then, $s(11k)=2s(k)$, absurd. If $k$ has 2 digits, then $s(101k)=2s(k)$, contradiction. If $k$ has 3 digits, then $s(1001k)=2s(k)$, again a contradiction. Then, we have that $k$ has at least 4 digits. If $k$ have exactly four digits (say $k=a_3 a_2 a_1 a_0$), we claim that $k=9999$. We know that $s(1001k)=s(k)$. Writing $1001k$ it as: $a_3- a_2- a_1- a_0 -0 -0-0$ $+0-0-0-a_3 -a_2 -a_1 -a_0$ we see that we must have 1 carrier after the sum $a_3+a_0$ (we are summing right to left), because if that's not the case, the sum of the digits of $1001k$ will be $2(a_3+a_2+a_1+a_0)>s(k)$, absurd. Also, if $a_1<9$, the sum of the digits of $1001k$ is at least $a_3+a_2+(a_1+1)+0+a_2+a_1+a_0>s(k)$, contradiction. So, $a_1=9$ and if $a_2<9$, the sum of digits of $1001k$ is at least $a_3+(a_2+1)+0+0+a_2+a_1+a_0>s(k)$, contradiction. So, $a_1=a_2=9$ and if $a_3<9$, the sum of digits of $1001k$ is at least $(a_3+1)+0+0+0+a_2+a_1+a_0>s(k)$, contradiction. Thus, $k=999a_0$, and notice that $a_0 \ne 0$, because if so, $1001k=9990999$, implying $s(1001k)=54=2s(k)$, contradiction. Also, $k$ is divisble by 9 (because $k \equiv s(k)=s(2k) \equiv 2k (mod. 9)$. So, we have $k=9999$, showing us that the minimum is at least 9999. Now we show that $k=9999$ solves the problem, by showing that $s(ak)=36$, for all $k=1,2,...,2014$ (actually, that works or every $a$ with at most 4 digits). Let $a=a_3a_2a_1a_0$ (we can have $a_3=0$), and first suppose that $a_0 \ne 0$. Then, we can write $9999a$ as $a_3 a_2 a_1 a_0 0000-0000 a_3 a_2 a_1 a_0 = a_3 a_2 a_1 (a_0-1)(9-a_3)(9-a_2)(9-a_1)(10-a_0)$ , and the sum of the digits is precisely 36. Now, if $a_0=0$, $s(ak)=s((a/10)k)=36$ if $a_1 \ne 0$. If $a_0=a_1=0$, $s(ak)=s((a/100)k)=36$, if $a_2=0$. If $a_0=a_1=a_2=0$, $s(ak)=s((a/1000)k)=36$, since $a_3 \ne 0$. Thus, $s(ak)=36$, for all $k=1,2,...,2014$, showing that $k=9999$ is solution, and its the minimum because of the preceding paragraph.

I think we can proceed in the following way First prove that if $k=9999$ required sum is $s(nk)$ is $36$ for all $n < 10001$ just as Davi Medeiros did .Now we will have to prove the minimality. Let $x$ be the minimum $k$ Observe that $x$ is divisible by $9$ since if not $s(x)$ is not a multiple of $9$ but $s(9x)$ is .Now consider $s(1111x)$ .Since $1111x$ is divisible by $9999$ and $x <10000 \Rightarrow s(1111x) = 36 \Rightarrow s(x)$ is $36, \Rightarrow x=9999$ since no number less than $9999$ has a digital sum greater than or equal to $36$.