In a company of people who know each other well, everyone either always tells the truth or always lies. Sitting around a round table, everyone claims that they are a truth-teller, but the person sitting $k$ seats to their right is a liar. How many people can sit around the table?
Let $n$ people sit at the table.

Let $t$ be the smallest integer - depending only on $n$ and $k$ - that indicates how many times $k$ steps must be taken around the table to return to the starting point. In such a round trip, a truth-teller is always followed by a liar, and a liar by a truth-teller, so the length of the cycle must necessarily be even. (Otherwise, there would be two "adjacent" truth-tellers or liars.)

We will show that if - for a given $n$ and $k$ - $t$ is even, then the appropriate arrangement is possible. Let's label the seats (or the people sitting in them) as follows. Pick one, let it be $\mathrm{I}$ (truth-teller), the one sitting $k$ places to the right is $\mathrm{H}$ (liar), the one $k$ places to the right of that is again $\mathrm{I}$, and so on; in $t$ steps we return to the starting point, and the last, $t$-th label was $\mathrm{H}$.

If there is a place/person that has not yet received a label, pick one such, and starting from it, perform a labeling similar to the previous one. In doing so, we cannot reach a place where we have already been in the previous segment, so another $t$ labels can be assigned; and so on.

What remains is to determine $t$. Since stepping in $k$-s around the $n$-person table we circle it over and over - exactly $t$ times - $t$ is the minimum of those integers for which $n \mid t \cdot k$, that is, $\frac{n}{(n, k)} \left\lvert\, t \cdot \frac{k}{(n, k)}\right.$. Since $\frac{n}{(n, k)}$ and $\frac{k}{(n, k)}$ are relatively prime to each other, $\left.\frac{n}{(n, k)} \right\rvert\, t$; the smallest such $t$ is $\frac{n}{(n, k)}$.

The solution to the problem is thus given by those values of $n$ for which $t=\frac{n}{(n, k)}$ is even, that is, $n$ is divisible by a higher power of 2 than $k$.