A hare and a tortoise run in the same direction, at constant but different speeds, around the base of a tall square tower. They start together at the same vertex, and the run ends when both return to the initial vertex simultaneously for the first time. Suppose the hare runs with speed 1, and the tortoise with speed less than 1. For what rational numbers $q{}$ is it true that, if the tortoise runs with speed $q{}$, the fraction of the entire run for which the tortoise can see the hare is also $q{}$?
1. Let the speed of the hare be \(1\) and the speed of the tortoise be \(q\), where \(q\) is a rational number less than \(1\). We need to find the rational numbers \(q\) such that the fraction of the entire run for which the tortoise can see the hare is also \(q\).

2. Suppose the tortoise runs with speed \(q = \frac{m}{n}\), where \(\gcd(m, n) = 1\). The hare and the tortoise start together at the same vertex of the square tower and return to the initial vertex simultaneously for the first time after \(4mn\) units of time, since the perimeter of the square is \(4\) and the least common multiple of their times to complete one lap is \(4mn\).

3. Define \(S(m, n)\) as the number of times the tortoise can see the hare during the \(4mn\) units of time. Specifically,
   \[
   S(m, n) := \#\left\{0 \le k \le 4mn - 1 : \lfloor k/m \rfloor \equiv \lfloor k/n \rfloor \pmod{4}\right\}.
   \]

4. The fraction of the time for which the tortoise can see the hare is given by:
   \[
   q = \frac{S(m, n)}{4mn}.
   \]

5. We need to calculate \(S(m, n)\) for different values of \(m\) and \(n\). The key cases are:
   \[
   S(m, n) = \begin{cases}
   mn & \text{if } 2 \mid mn \\
   mn + 3 & \text{if } m \equiv n \in \{1, 3\} \pmod{4} \\
   mn + 1 & \text{if } m \equiv -n \in \{1, 3\} \pmod{4}
   \end{cases}
   \]

6. By substituting these values into the equation \(q = \frac{S(m, n)}{4mn}\), we get:
   \[
   q = \frac{mn}{4mn} = \frac{1}{4}, \quad q = \frac{mn + 3}{4mn} = \frac{3}{11}, \quad q = \frac{mn + 1}{4mn} = \frac{1}{3}.
   \]

7. Therefore, the rational numbers \(q\) for which the fraction of the entire run for which the tortoise can see the hare is also \(q\) are:
   \[
   q = \frac{1}{4}, \quad q = \frac{3}{11}, \quad q = \frac{1}{3}.
   \]

The final answer is \(\boxed{q = \frac{1}{4}, \frac{3}{11}, \frac{1}{3}}\).