Will there always be 256 collisions for the output of the MD5 hashsum function for 17-byte inputs?
Your math is wrong — not the numerical calculation, but your interpretation of it. There are $256^{17}$ possible inputs and $256^{16}$ possible outputs. On average, there are $256$ inputs for each output. But there are no guarantees that this is the case for all outputs: it's in fact overwhelmingly likely that some outputs have more and others have fewer. For example, we don't even know whether every possible output is in fact the MD5 of a message of any length.

What you can state for sure is that there exists at least one 256-bit string $H$ such that there are at least 256 17-byte messages $M_{1}, \ldots, M_{256}$ such that for all $i$, $\mathrm{MD5}(M_i) = H$. Since we don't know any MD5 collision with a 17-byte output, we are unable to come up with an actual value for $M_i$ or $H$, we just know that it exists by the pigeonhole principle.

See also


Is SHA-256 a one way permutation when restricting the domain to {0,1}^256?
Is it theoretically possible to construct a string that contains its own hash value?
What is the MD5 collision with the smallest input values?