“Solve” means “solve the halting problem for” (i.e. prove whether it will halt or not), and “exhaustively” means “for all possible Turing machines with a given number of states”
But intuitively a Turning machine, no matter how small number of states, can't be exhaustively solved, since there are infinite number of possilbe strings on the tape?
Proving properties hold over sets of infinite cardinality is fairly routine in mathematics.
We know that not every Turing Machine can be “solved”, because this would give us an oracle for HALT, but there are machines that can be.
A trivial example is a TM that halts immediately, regardless of the input; a less trivial example is one that reads an arbitrarily large integer input on the tape and outputs its factorisation (or, if we don’t want to consider output, decides whether an input pair is formed of an integer and it’s factorisation).
Given that the set of Turing Machines with a certain number of states (or less) is finite for a given alphabet, it’s “just” a case of solving that finite set.
EDIT: Note that as the sibling comment points out, BB is usually considered to start with an empty tape, but fundamentally, proving that a machine always halts regardless of their starting tape is not intrinsically impossible as you suggest.
Generally, when we talk about the halting problem in the Busy Beaver context, we're talking about the halting problem starting from an empty tape. However, some authors have analyzed the behavior of Turing machines on certain classes of infinite tapes. For instance, in [0], the authors surveyed the behavior of TMs starting on a tape with an arbitrary word in the middle, and an infinitely repeating word on one or both sides; one finding is that no 2-state 2-symbol TM is a universal Turing machine, for any of these input tapes.
