Not sure if I screwed up? If $p \neq 2,5, n=10^{(p-1)k}$ works. If $p=2, n=111...111$ works (odd number of 1s). If $p=5, n=555...555$ works.

yes it works

Good problem

ManuelKahayon wrote: Let \(S(n)\) denote the sum of the digits of the base-10 representation of an natural number \(n\). For example. \(S(2017) = 2+0+1+7 = 10\). Prove that for all primes \(p\), there exists infinitely many \(n\) which satisfy \(S(n) \equiv n \mod p\). Let $n=1$, ...

This was really simple ,and there should be many constructions possible. Just take $n=1111..111$ with $k$ ones such that $p(p-1)|k$ and we are done.