$$n$$ is a natural number and $$n \le 2026$$. If $$n$$ divides $$20^n + 17k$$, for some natural number $$k$$, then find the number of such $$n$$.

If $$17 \mid n$$, then 17 would have to divide $$20^n + 17k$$ and hence $$20^n$$, which is impossible since $$20 = 2^2 \times 5$$; so no multiple of 17 works. If $$\gcd(17, n) = 1$$, then as $$k$$ runs through $$1, 2, \ldots, n$$ the numbers $$17k$$ leave every remainder $$0, 1, \ldots, n - 1$$ on division by $$n$$, so we can choose $$k$$ with $$17k \equiv -20^n \pmod{n}$$, which makes $$n \mid 20^n + 17k$$. So every $$n \le 2026$$ works except the multiples of 17, of which there are $$\left\lfloor \frac{2026}{17} \right\rfloor = 119$$. The required count is $$2026 - 119 = 1907$$.