Solution:

We indicate all lengths in meters and all times in seconds, omitting the units of measure.
The possible triples of speeds are $\left(v_{s}, v_{p}, v_{d}\right)=(10,12,15),(9,12,18),(8,12,24)$ and $(7,12,42)$.

Let us first consider a completely flat circuit of length 600. Hypothesis (c) then gives $50=\frac{600}{v_{p}}$, from which $v_{p}=\frac{600}{50}=12$.

Let us then consider a generic circuit of length 600, in which the uphill section has length $s$, the flat section $p$ and the downhill section $d$.
By construction $s+p+d=600$; moreover, in order for the circuit to close, the elevation gain given by the uphill sections must be the same as that given by the downhill sections: since the slopes are the same for all the non-flat sections, there must be as much uphill as downhill, that is $s=d$.

The flat section then has length $p=600-s-d=600-2s$, and the time taken by the car to travel the circuit is $\frac{s}{v_{s}}+\frac{p}{v_{p}}+\frac{d}{v_{d}}=\frac{s}{v_{s}}+\frac{600-2s}{12}+\frac{s}{v_{d}}=s\left(\frac{1}{v_{s}}+\frac{1}{v_{d}}-\frac{1}{6}\right)+\frac{600}{12}$, which is equal to $50=\frac{600}{12}$ if and only if $s\left(\frac{1}{v_{s}}+\frac{1}{v_{d}}-\frac{1}{6}\right)=0$.

The speeds then satisfy condition (c) of the statement if and only if the previous relation holds for any closed circuit whatsoever, that is, for any choice of $s$ (between 0 and 300).

A necessary and sufficient condition on the speeds is therefore $\frac{1}{v_{s}}+\frac{1}{v_{d}}=\frac{1}{6}$; $v_{s}$ and $v_{d}$ are both different from zero, so we can rewrite this equation as $v_{s} v_{d}=6\left(v_{s}+v_{d}\right)$, that is $\left(v_{s}-6\right)\left(v_{d}-6\right)=36$.

$v_{d}$ is an integer greater than $v_{p}=12$, so $v_{d}-6$ is a divisor of 36 strictly greater than 6, that is it is one of 9, 12, 18 or 36 (which correspond respectively to the values $15, 18, 24, 42$ for $v_{d}$).

Solving the equation for $v_{s}$ we finally obtain $v_{s}=6+\frac{36}{v_{d}-6}$, that is $v_{s}=10,9,8,7$.

Solution:

Once we have obtained the equation $\frac{1}{v_{s}}+\frac{1}{v_{d}}=\frac{1}{6}$ it suffices to observe that $v_{s}<v_{p}=12$, and on the other hand we cannot have $v_{s} \leq 6$, because otherwise $\frac{1}{v_{s}} \geq \frac{1}{6}$, from which $\frac{1}{v_{d}} \leq 0$, which is clearly absurd. It follows that the only possibilities for $v_{s}$ are $7,8,9,10,11$, and it is easy to check which of these lead to integer values for $v_{d}$.