I'll reformulate the question in light of recent addition. It asks: are modern cryptographic hash functions applied to consecutive integers periodical?. Let's assume agreement on how we hash an integer, even though two practitioners asked the common method will come with 10 different¹.
An answer argues that no, on the ground that for a hash function as in the question, $\operatorname{Hash}(n)=\operatorname{Hash}(n+k)$ with $0<k$, which is a collision, and goes against one of the design property of cryptographic hashes since their inception: collision-resistance. But, as the OP rightly commented, that reasoning falls short: a period could exist but be so large that it can't be found.
That same no conclusion can be reached by another line of reasoning: a modern hash attempts to be, and is often modeled as, a random function from $\{0,1\}^*$ (the set of finite bitstrings, which is infinite) to $\{0,1\}^h$ (the set of $h$-bit bistrings). Under that model, regardless of the mapping of integer to bitstring, the series of outputs of the hash with input the consecutive integers is a set of random bitsrings in $\{0,1\}^h$. It can be rigorously defined and computed the probability that such series is periodic, and that's zero².
But SHA-256 falls short of the above model: there are "only" $2^{(2^{64})}-1$ valid inputs to SHA-256, thus no matter how we uniquely map integers to bitstrings, the series has a finite number of terms $t$, with $t<2^{64}$. Hence the "all positive integers $a$" in the question must be somewhat restricted so that $\operatorname{SHA-256}(n+b+a\,k)$ remains well defined. And when we make that restriction, yes the condition stated in the question is met: $n=t-1$, $k=1$ (allowing only $b=0$) match it, for there remains no valid value of $a$ to contradict periodicity.
But SHA-256 was intended only as example, and SHA-3 also qualifies as a modern hash. AFAIK it's input set is $\{0,1\}^*$, thus the above argument fails; and SHA-3 is deterministic, processes it's input sequentially by finite blocks and has a finite state, invalidating the random function model. This reopens the question. I believe it's not periodical for any natural mapping of integers to bitstring, including if we stretch that to use base $2^b$ where $b$ is the input block size, and special-case the last block to make it constant; but I have no proof.
¹ I'm not joking about the 10 kinds of people: those who understand binary, and the others. I'm considering:
- bases, with 2, 10, 16, 64, 256 (raw bytes), 232, or 264 the most common;
- endianness, which can be big, little, or blended in a variety of ways;
- and worse, standards (obligatory XKCD). One of several common in crypto is ASN.1 DER, which codes the integer 0 as the three bytes
02 01 00, $2^{1015}-1$ as 129 bytes, and the next integer with two extra bytes ($2^{1015}$ is coded as 131 bytes).
² Sketch: for any periodicity defined by $k$ and the minimum $n$, each additional value starting $n+2k$ only has probability $2^{-h}$ to match the periodicity. If we analytically compute the probability of that property after $x$ terms, and sum these over all the $k,n$ meaningful for $x$ terms, what we obtain goes to zero when $x$ goes to infinity.
