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17k/66 and 13k/105 must reduce to fractions with a denominator that only consists of powers of 2 or 5.
For example, some fractions with terminating decimals are
1/2 = 0.5
1/4 = 1/2² = 0.25
1/5 = 0.2
1/8 = 1/2³ = 0.125
1/10 = 1/(2•5) = 0.1
1/16 = 1/2⁴ = 0.0625
and so on, while some fractions with non-terminating decimals have denominators that include factors other than 2 or 5, like
1/3 = 0.333…
1/6 = 1/(2•3) = 0.1666…
1/7 = 0.142857…
1/9 = 1/3² = 0.111…
1/11 = 0.09…
1/12 = 1/(2²•3) = 0.8333…
etc.
Since 66 = 2•3•11, we need 17k to have a factorization that eliminates both 3 and 11.
Similarly, since 105 = 3•5•7, we need 13k to eliminate the factors of 3 and 7.
In other words, 17k must be divisible by both 3 and 11, and 13k must be divisible by both 3 and 7. But 13 and 17 are both prime, so it's just k that must be divisible by 3, 7, and 11. These three numbers are relatively prime, so the least positive k that meets the conditions is LCM(2, 7, 11) = 231, and thus k can be any multiple of 231.
If you're familiar with modular arithmetic, this is the same as solving for k such that
13k ≡ 0 (mod 3)
17k ≡ 0 (mod 3)
17k ≡ 0 (mod 7)
13k ≡ 0 (mod 11)
and the Chinese remainder theorem says that k = 231n solves the system of congruences, where n is any integer.
Now it's just a matter of finding the smallest multiple of 231 that's larger than 2000, which easily done by observing
2000 = 8•231 + 152
and so k = 9•231 = 2079.