= Solution
This is <linked file allocation>. A <file header> holds the first and last <disk block numbers>, and the successor field in each block extends the chain. The header therefore does not impose a fixed small maximum number of blocks per file.
To state a numerical bound, suppose a physical block has $B$ bytes, the successor-or-final-count field occupies $h$ bytes, and $N$ distinct data blocks are both addressable and available. A full block then holds $C=B-h$ payload bytes. If the final count can represent every value up to $C$,
$$
\boxed{L\leq NC=N(B-h).}
$$
A smaller <computer file> can use an arbitrary number of full payload blocks followed by a partial block, so allocation need not be contiguous. With a $p$-bit <disk block number>, $N\leq2^p$, reduced by reserved addresses and blocks consumed by other data and metadata. This gives the address-width bound as well as the actual device-space bound.
There is no positive minimum payload size: an empty <computer file> can be represented by an empty-chain convention or a zero-count terminal block. The question does not specify which representation is used or the relevant field widths, so it does not determine an absolute byte limit or a particular empty-file encoding. Its description presupposes a convention that distinguishes a successor value from a terminal byte count.
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