There can't be, since KC isn't computable (could be mistaken on that in itself precluding a bounded error).

KC may not be uncomputable *in general*, but I'm pretty sure that doesn't preclude all possible proofs or constructions*, and it seems odd to say that there is no upper bounds when we have what sure look like such things.

* I have just invented a Scheme-like programming language in which all tokens except numbers are 2 alphanumeric characters long; what is the Kolmogorov complexity of the bitstring '1'? 'Who knows, KC is uncomputable -' Bzzt! The answer is that the KC is exactly 1, since the shortest program which emits the bitstring '1' is a program consisting of the constant '1' which evaluates to '1', which as you can see, is indeed of length 1, and all other programs emitting the bitstring '1' are longer by definition.

Or if you don't like that example, consider taking your favorite language and enumerating all possible programs in length order; consider the outputs of the programs you've just enumerated when evaluated up to the highest available Busy Beaver in steps (to avoid nontermination issues), and the respective lengths of outputs - have you not found the KC for those outputs? If you run gzip over the outputs to get estimated KCs, are those not upper bounds on the actual KCs you proved?

There is no such thing as "the shortest program for which the halting property is uncomputable". That property is computable trivially for every possible program. What's uncomputable is always the halting problem for an infinity of programs using one common algorithm.

It is also easy to make up artificial languages in which Kolmogorov complexity is computable for an infinite subset of all possible strings.

You were probably thinking of something else: that there exists a constant L, which depends on the language and a proof system T, such that it's not possible to prove in T that any string has Kolmogorov complexity larger than L. That is true. In particular, this means that there's a limit to lower bounds we can establish, although we don't know what that limit is.

*2 points [-]The halting property is semi-decidable: if a program halts, then you can always trivially prove that it halts, you just need run it. If a program does not halt, then sometimes you can prove that it doesn't halt, and sometimes you can't prove anything.

For any programming language, there exist a length such that you can compute the halting property for all programs shorter than that. At length L_max, there will be program Bad_0, which:

You can never prove that any string S has Kolmogorov complexity K if K > L_max, as it would imply that you proved that Bad_0 doesn't halt, or at least doesn't emit any symbol which is not a prefix of S.

Since there are only finitely many strings with complexity up to L_max, we can only compute Kolmogorov complexity, for finitely many strings.

If the language is Turing-complete I don't think this is possible. If you think that an arbitrary string S has complexity K, how can you prove that there exist no program shorter than K that computes S?