Capacity of a discrete memoryless channel
This answer assumes that $Y = X + Z \bmod 2I$, which seems more reasonable than the current $Y = X + Z \bmod I$, which would mean that $X$ is effectively chosen from an alphabet of size $I$ rather than $2I$.

The capacity is $\log I$.

For the upper bound,
$$
I(X;Y) = H(Y) - H(Y|X) = H(Y) - H(Z) = H(Y) - 1 \leq \log(2I) - 1 = \log I.
$$

For the lower bound, if $X$ is chosen uniformly among the $I$ values $1,3,\ldots,2I-1$ then you can recover $X$ from $Y$ (without any error!), and so $I(X;Y) = H(X) = \log I$.