Suppose we have a display with the sequence of $n$ zeros and ones on it. We are allowed to alter no more than $m\leq n$ of it. After that our friend comes to the display to get $m$ bits of messege. He knows $m$, but he doesn't know which bits we were allowed to change. Our friend and we can discuss beforehand our strategy. How should we act?
I can only solve it for $m=1$ (+ a trivial solution for $n=m$). By altering just one bit we can make the modulo-2 sum of all bits as we want, and thus transmit one bit of information.
By altering $m$ bits we can make the modulo-$(m+1)$ sum of all bits as we want, and thus transmit the number from $0$ to $m$, i.e. transmit $\log_2(m+1)$ bits of information. But that's good enough only for $m=1$...
This is possible for $m = n-1$ as well, but not otherwise. In order to be decodable, for each input message we must have a collection of $2^{n-m}$ output messages, at least one of which is attainable given any set of fixed bits. There are $2^{n-m}\cdot {{n} \choose {m}}$ such initial configurations of fixed bits, so each output message must correspond to a unique set of ${{n} \choose {m}}$ initial configurations.
Note that two output messages correspond to disjoint sets of initial configurations if and only if they have fewer than $n-m$ bits in common. To see this, note that two output messages can come from the same initial configuration if they agree on the fixed bits of that configuration. So if two output strings agree on $n-m$ bits, there is some initial configuration that allows both output strings. Alternately, if there is no set of $n-m$ bits on which they agree, then no initial configuration will allow constructing both.
It is not in general possible to produce $2^{n-m}$ $n$-bit strings such that every pair has at most $n-m-1$ bits in common. For instance, take $m=2$ and $n=4$. Then without loss of generality say the first string is $0000$. Taking the second string as one with at least three $1$s, any third string has either two ones or two zeros. In general, it seems that if $n > m+1$ such a subdivision is not possible. (This seems to have something to do with the fact that the Hamming distance is a metric, and the diameter of the space of $n$-bit strings is $n$.)
However, to go back to the case $m=n-1$, we can do this because we only need to have $2$ output messages for each input, which we can do by picking a complementary pair of outputs for each input.