Solution:

The answer is $\mathbf{( C )}$. We know that Zanobi is slower than Veronica. In particular, there exists a factor $k>1$ such that $v_{V}=k v_{Z}$, where $v_{V}$ and $v_{Z}$ are the speeds of Veronica and Zanobi, respectively.

Veronica completes 28 lengths in time $T$. We can compute the number $L_{Z}$ of lengths covered by Zanobi as follows: Veronica's speed is $v_{V}=28 / T$, Zanobi's is $v_{Z}=L_{Z} / T$. But then
$$L_{Z}=T v_{Z}=\frac{28}{v_{V}} v_{Z}=\frac{28}{k}$$
At the end of the 28 lengths, Veronica is on the same side from which she started and, since Zanobi is level with her, he too must have completed an integer and even number of lengths. In order for $k$ to be a valid value, therefore, we must have $k=28 / L_{Z}>1$, where $L_{Z}$ is an even number greater than zero and less than 28.

We can repeat the same reasoning when Veronica completes 70 lengths: we must require that $70 / k$ be an even integer, but
$$70 / k=L_{Z} \cdot 70 / 28=L_{Z} \cdot 5 / 2=5 \cdot\left(L_{Z} / 2\right)$$
which is always an integer, and is even if and only if $L_{Z}$ is a multiple of 4. The only valid values of $L_{Z}$ and $k$ are therefore $L_{Z} \in\{4,8,12,16,20,24\}$, from which $k=28 / L_{Z} \in\{7,7 / 2,7 / 3,7 / 4,7 / 5,7 / 6\}$. These correspond to the following possible values for the number $m$ of lengths completed by Zanobi:
$$m=70 / k \in\{10,20,30,40,50,60\} .$$