Murashkin M.V.

In a racing tournament, there are 12 stages and $n$ participants. After each stage, all participants receive points $a_{k}$ depending on the place $k$ they occupy (the numbers $a_{k}$ are natural, and $a_{1}>a_{2}>\ldots>a_{n}$). For what smallest $n$ can the tournament organizer choose the numbers $a_{1}, \ldots, a_{n}$ so that after the penultimate stage, regardless of the possible distribution of places, at least two participants have a chance to take the first place.
Evaluation. Suppose there are no more than 12 participants.

Let one of the participants (call him A) win all 11 stages, and each of the remaining participants at least once took the last place.

Then participant $A$ after 12 stages will score no less than $11 a_{1} + a_{n}$ points, while each of the remaining participants will score no more than $a_{n} + 10 a_{2} + a_{1}$ points, which is less than

$11 a_{1} + a_{n}$

Therefore, only $A$ can take the first place. Hence, to meet the condition, there must be at least 13 participants.

Example. Suppose there are 13 participants, then the organizer can choose the numbers $a_{1}=1011, a_{2}=1010, \ldots, a_{12}=1000, a_{13}=1$.

Consider the participant who scored the most points after 11 stages (call him $A$). Then among the remaining 12 participants, there will be at least one who did not take the last place on any stage (call him $B$). Note that $A$ scored no more than $1011 \cdot 11 = 11121$ points, while $B$ scored no less than $1000 \cdot 11 = 11000$ points. That is, after 11 stages, $A$ is ahead of $B$ by no more than 121 points.

Obviously, after 11 stages, $A$ has a chance to take the first place. However, if on the 12th stage $A$ takes the last place, then $B$ will score at least 999 points more than $A$ on the 12th stage and overtake him.

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