The case $r = 2$ is clear: Any cycle of even length works. In the other cases, define a sequence of graphs $G_r$, $r \ge 3$, as follows. The graph $G_3$ is a cycle of just 7 edges. (Any larger odd number would do.) When $G_r$ is defined, with $n_r$ vertices say, construct $G_{r+1}$ as follows. Consider

$$\binom{rn_r - r + 1}{n_r}$$
disjoint copies of $G_r$. Adjoin $rn_r - r + 1$ extra vertices. Set up a one-to-one correspondence between the copies of $G_r$ and the $n_r$-element sets of extra vertices. Join each copy of $G_r$ to the members of the corresponding $n_r$-element set of extra vertices by $n_r$ disjoint new edges (no two have a common end). The resulting graph is $G_{r+1}$.

Clearly, $G_3$ is 3-chromatic. If $r \ge 3$ and $G_{r+1}$ has a coloring $C$ in $r$ or fewer colors, then some $n_r$ of the extra vertices in $G_{r+1}$ must share the same color in $C$, so the corresponding copy of $G_r$ must be colored in $r-1$ or fewer colors. It follows, by descending induction, that $G_3$ must be colored in 2 or fewer colors which is, of course, impossible. Consequently, no $G_r$ can be colored in less than $r$ colors.
This does not prove that $G_r$ is $r$-chromatic, but if it is not, deletion of some monochromatic classes of vertices together with their incident edges yields one such.

The construction ensures that no graph $G_r$ has a cycle of less than 6 edges.