Let forbidden=indecent. Evaluation: obviously, $(a,a)$ and one of $(a,b),(b,a)$ are forbidden (try words $aaa...$ and $abab...$) ($a,b$ are letters). Then at least $n+n(n-1)/2=n(n+1)/2$ is forbidden in total. Example: let's arrange the letters $1<2<...<n$ and forbid $(i,i+k)$ for non-negative $k$. Then the letter numbers in the word are strictly reduced, so the length of the words is no more than $n$. So, the answer is $n(n+1)/2$.

I kinda did not like the formulation in this problem since it can be interpreted in two ways (several students asked about it during the competition), though in one of the interpretations the problem is even more trivial. 1) If the answer is $k$, then you are expected to show that no choice of $k-1$ syllables forbids all infinite words and that for some choice of $k$ syllables all infinite words are forbidden. 2) A different way one could interpret the problem statement instead is that if the answer is $k$, then you are expected to show that any choice of $k$ syllables forbids all infinite words and some choice of $k-1$ syllables does not forbid all infinite words. (Answer: $n^2$, as for $n^2-1$ forbidding everything except $(1,1)$ still allows the word of 1s.)

Let $G$ be a directed graph with letters as vertices and $XY$ exists iff $XY$ is not indecent. This graph can't contain any cycles, so it can have at most $\frac{n(n-1)}{2}$ edges out of the possible $n^2$ (cus $XX$ can exist), so the total number of removed edges is at least $\frac{n(n+1)}{2}$, and you can construct this by ordering vertices and connecting the ones with smaller indices to the ones with bigger indices.

Answer is $\frac{n^2+n}{2}$. Necessary because if less, then in $n$ by $n$ table representing syllables of length two either a syllable on diagonal is indecent, and we just spam it, or a pair of cells symmetric with respect to diagonal are both indecent, and we just alternate. Sufficient because if in the same table, we make all syllables on the diagonal and below it indecent, then after any syllable is placed first, there is no way to add more syllables.

Bit easy for IZHO p1

iflugkiwmu wrote: Answer is $\frac{n^2-n}{2}$. Necessary because if less, then in $n$ by $n$ table representing syllables of length two either a syllable on diagonal is indecent, and we just spam it, or a pair of cells symmetric with respect to diagonal are both indecent, and we just alternate. Sufficient because if in the same table, we make all syllables on the diagonal and below it indecent, then after any syllable is placed first, there is no way to add more syllables. @above Correct your answer

Too easy！

Answer: $\frac{1}{2}(n^2+n)$ Construction: Let the letters be $l_1,l_2,\dots,l_n$. A syllable $l_il_j$ is considered idecent if $i\leq j$. In any word the indicies of the letters must be strictly decreasing, thus the word is finite. Bound: If $A$ and $B$ are letters then $AA$, $BB$, and one of $AB$ or $BA$ must be indecent. This leads to the desired bound.