Twelve candidates for the position of mayor are participating in a televised debate. After a while, one of them declares, "Up until now, we have lied once." A second then says, "Now, it's twice." A third exclaims, "Three times, now," and so on until the twelfth, who asserts that before him, there were twelve lies. The host then stops the discussion. Knowing that at least one of the candidates correctly announced how many times they had lied before their turn to speak, determine how many candidates lied in total.
We denote $C_{1}, \ldots, C_{12}$ as the candidates in the order of their speaking. Since at least one of them told the truth, we can consider the smallest integer $k$ such that $C_{k}$ told the truth (i.e., $C_{k}$ is the first to tell the truth). By the minimality of $k$, all those who spoke before him lied. Furthermore, since $C_{k}$ told the truth, there were exactly $k$ lies before his turn, and thus exactly $k$ lies after. Therefore, $C_{k+1}$ lied by saying there were $k+1$ lies before his turn, and there were indeed $k+1$ lies after $C_{k+1}$ spoke. Consequently, $C_{k+2}$ lied by saying there were $k+2$ lies before his turn, and there were indeed $k+2$ lies after $C_{k+2}$ spoke, and so on. Thus, all candidates who spoke after $C_{k}$ lied.

And, finally, only $C_{k}$ told the truth. Therefore, there were 11 liars.

Remark. In fact, we can specify the situation a bit more. Indeed, if $k \geq 2$, then with the above notation, we know that $C_{k}$ told the truth and that there were exactly $k$ lies before his turn. Taking into account the lie of $C_{k-1}$, it means there were exactly $k-1$ lies before $C_{k-1}$ spoke. But then, $C_{k-1}$ told the truth, which contradicts the minimality of $k$. Therefore, we must have $k=1$, which means that the first candidate to speak is the only one who told the truth.