Terminal Lemma

Where k=1, s(ka)+s(kb) must be 19 or more. Where b is 18, values from k=1 to k=18 proceed from 18 to 1, as it goes from a difference of 1 to multiples of that difference. Conveniently, that also means k increases at the rate s(k 18) decreases. The sum of the two must always be 19. That means that s(ka) must be at least 1 for the condition, k+s(ka)+s(kb)≥20 to be met. As s(x) will only be 0 if x is a multiple of 19, and x will only be a multiple of 19 if 19 is a factor, as 19 is prime. Obviously, if neither a or b can be a factor, a b cannot be divisible by 19. As such, all solutions where b=18 are solutions. There are 18 of these, from a=1 to a=18. However, there are more solutions. The only other conditions under which this solution holds are where a+b=19. This is because the addition rule of modular arithemtic states that (a%b+c%b)%b=(a+b)%c. For this equation to be true where b is not only 18, all values of s(s(ka)+s(kb)) must be 0. The only way that this exceeds 20 when added to k is when it is 19, the sum of two residues of 19 cannot be 19*2 or greater. s(ka)+s(kb) will only always be 19 where a+b=19. There are 8 such solutions where a<b. 8+18=26

(Disclaimer: this solution needs some work / further thought)

First off, for when $k = 1$ , clearly that makes a restriction $a + b \geq 19$ .

The pairs $(a, 19 - a)$ and $(a, 18)$ both work (these are also the only ones, though I do not have a formal argument yet as to why the others don't work -- hence the disclaimer).

$(a, 19-a)$ works because $s(ka) \equiv -s(kb) \pmod {19}$ , thus $s(ka) \pmod {19} + s(kb) \pmod {19} = 19$ . Since $k \geq 1$ , then it follows that $k + s(ka) + s(kb) \geq 20$ .

$(a, 18)$ works because $k = k \pmod {19}$ and $s(18k) = -k \pmod {19}$ , thus $k \pmod {19} + s(18k) \pmod {19} = 19$ . Since $19$ is prime, then $s(ka) \not\equiv 0$ for $1 \leq k, a \leq 18$ . Thus, it follows that $k + s(ka) + s(kb) \geq 20$ .

Adding up these two cases, the total number is $26$ .