What you are describing is a modification of the $q$-ary symmetric channel. As pointed out by Yuchiro in a previous answer, your problem is trivial when $q = 2$, meaning that if we have a binary alphabet, then knowing the position of the errors actually tells you what are the correct bits.
Suppose you have an alphabet $\{0,1,2\}$, of size $3$. The way you described you problem, if the symbol $0$ is sent, we can receive either $0$ or $1',2'$, i.e., we know whether the received symbol is wrong. The same happens for $1$ and $2$. Thus, the channel for your model can be described by the following graph: ![Channel graph][1]
(I am supposing that the transition probabilities are symmetric). Considering $y$ the output and $x$ the input, the transition matrix is $$P_{ij} = P(y = j |x = i) = \left( \begin{array}{cccccc} p_1 & 0 & 0 & 0 & p_2 & p_2 \\ 0 & p_1 & 0 & p_2 & 0 & p_2 \\ 0 & 0 & p_1 & p_2 & p_2 & 0 \\ \end{array} \right).$$ Maximizing mutual information gives you the capacity. I made a quick calculation and may be wrong, but I got that the capacity is
$$C = p_1 \log 3 + 2 p_2(\log 3 - 1) \mbox{ bits / channel use }$$. This makes a lot of sense, since if $p_1 = 1$ this is $\log 3$, a perfect channel, and if $p_1 = 0$, this is $\log 3 -1$, the capacity of a symmetric channel consisting only of the last three symbols $0', 1', 2'$ as output.
Now you should be able to generalize this for any $q$ with a little extra effort.
Thus, the answer to your question if this is "classic" is "yes", in the sense that you can achieve capacity doing Shannon type strategies (random coding, etc..) or even more modern stuff (e.g, polar codes). On the other hand, I don't think nobody considered the construction of practical codes for this specific channel (to the best of my knowledge). [1]: https://i.sstatic.net/lK1dj.png
