Summing the digits

Since $18k$ is divisible by $9$ , this means that $S(18k)$ must be a multiple of $9$ . For the smallest $k$ such that $S(18k_{m}) > 18$ , means that $S(18k_{m}) = 27$ .

That smallest $n$ such that $S(n)=27$ is $n = 999$ . Since $18k_m$ is even, $18k_m$ must be at least a $4$ -digit integer. The smallest $4$ -digit even integer such that $S(n)=27$ is $n=1998$ and since $1998= 111\times 18$ is divisible by $18$ , then $k_m$ must be $\boxed{111}$ .

Python 2.7:

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def S(k):
    return sum([int(i) for i in str(k)])
def goal(k):
    return S(18*k) > 18
k = 0
while not goal(k):
    k += 1
print "Answer:", k

$\text{A number with two digits will have maximum sum of the digits as} \\ \text {9+9=18. ..So k must at lest be three digit. Now, for n from 1 to 10,} \\ \text{18 * n the total of digits is 9.}\\ \text{While for n=11, 18 * 11=198, and 100 * 18=1800.}\\ \text {So (11 +100) * 18=198+1800=1998, sum of digits >18.}\\So ~ k=~~~~~\Large \color{#D61F06}{111}$