6. 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 laps in time $T$. We can calculate the number $L_{Z}$ of laps covered by Zanobi as follows: Veronica's speed is $v_{V}=28 / T$, and Zanobi's speed is $v_{Z}=L_{Z} / T$. Therefore,

$$
L_{Z}=T v_{Z}=\frac{28}{v_{V}} v_{Z}=\frac{28}{k}
$$

At the end of the 28 laps, Veronica is on the same bank from which she started, and since Zanobi is at the same height, he must have completed an even and integer number of laps. For $k$ to be a valid value, therefore, it must be $k=28 / L_{Z}>1$, where $L_{Z}$ is an even number greater than zero and less than 28.

We can repeat the same reasoning when Veronica completes 70 laps: we must ensure that $70 / k$ is even and integer, but

$$
70 / k=L_{Z} \cdot 70 / 28=L_{Z} \cdot 5 / 2=5 \cdot\left(L_{Z} / 2\right)
$$

which is always 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 laps made by Zanobi:

$$
m=70 / k \in\{10,20,30,40,50,60\} .
$$
6. Zanobi and Veronica go to the swimming pool together and start swimming back and forth at constant but different speeds, each in their own lane, starting from the same side of the pool. Veronica notices that, when she completes 28 laps (i.e., finishes swimming the length of the pool 28 times), Zanobi is next to her. As soon as she completes 70 laps, Veronica stops swimming and gets out of the water; at the same moment, Zanobi also reaches the edge of the pool, stops swimming, and gets out of the water next to Veronica. Zanobi, who is slower than Veronica, has swum $m$ laps; how many different values can $m$ take?
(A) 1
(B) 3
(C) 6
(D) 35
(E) 69