In an alphabet of $n$ letters, is $syllable$ is any ordered pair of two (not necessarily distinct) letters. Some syllables are considered $indecent$. A $word$ is any sequence, finite or infinite, of letters, that does not contain indecent syllables. Find the least possible number of  indecent syllables for which infinite words do not exist.
1. **Understanding the problem**: We need to find the minimum number of indecent syllables such that no infinite word can be formed. A syllable is an ordered pair of letters, and a word is a sequence of letters that does not contain any indecent syllables.

2. **Possible solution**: We are given that $\binom{n+1}{2}$ is a possible number of indecent syllables. Let's analyze why this is the case.

3. **Strictly decreasing sequences**: If we forbid all syllables $(i, j)$ with $i \le j$, then any word must be strictly decreasing. Since there are only $n$ letters, any strictly decreasing sequence must be finite. This gives us $\binom{n+1}{2}$ indecent syllables because:
   \[
   \binom{n+1}{2} = \frac{(n+1)n}{2}
   \]
   This counts all pairs $(i, j)$ where $i \le j$.

4. **Necessary conditions**: We need to ensure that no infinite word can be formed:
   - For any letter $i$, we must forbid the syllable $(i, i)$ to prevent the infinite repetition $i, i, i, \ldots$.
   - For any two distinct letters $i$ and $j$, we must forbid at least one of $(i, j)$ or $(j, i)$ to prevent the infinite alternation $i, j, i, j, \ldots$.

5. **Counting the necessary indecent syllables**:
   - There are $n$ syllables of the form $(i, i)$.
   - For each pair of distinct letters $(i, j)$, we need to forbid at least one of $(i, j)$ or $(j, i)$. There are $\binom{n}{2}$ such pairs, and each pair contributes at least one indecent syllable.

6. **Total number of indecent syllables**:
   \[
   n + \binom{n}{2} = n + \frac{n(n-1)}{2} = \frac{2n + n^2 - n}{2} = \frac{n^2 + n}{2} = \binom{n+1}{2}
   \]
   This matches the given solution.

7. **Conclusion**: The least possible number of indecent syllables for which infinite words do not exist is $\binom{n+1}{2}$.

The final answer is $\boxed{\binom{n+1}{2}}$