The 34th question: Given an integer $n \geq 4$, find the smallest $\lambda(n)$ such that for any non-negative real numbers $a_{1}, a_{2}, \ldots, a_{n}$ (indices taken modulo $n$), the inequality $\sum_{i=1}^{n}\left\{a_{i}\right\} a_{i+1} \leq \lambda(n)$ always holds.
The 34th question,
Solution: We first prove a lemma: Let the integer $\mathrm{n} \geq 4$, real numbers $r_{i} \in[0,1](1 \leq \mathrm{i} \leq \mathrm{n})$, satisfy $\sum_{i=1}^{n} r_{i}=r$, where $r \in \mathbb{N}$, and $r \leq n-1$, then $\sum_{i=1}^{n} r_{i} r_{i+1} \leq r-\frac{3}{4}$.
Proof of the lemma: If $\mathrm{r}=0$, the conclusion is obviously true. Now assume $\mathrm{r} \in \mathbb{Z}^{+}$.
Let $f\left(r_{1}, r_{2}, \ldots, r_{n}\right)=\sum_{i=1}^{n} r_{i} r_{i+1}$, then $f$ is a continuous function defined on a bounded closed set, so there exists a maximum value $S$.

Without loss of generality, assume that $f$ achieves $S$ at $\left(r_{1}, r_{2}, \ldots, r_{n}\right)$, and that $r_{1}, r_{2}, \ldots, r_{n}$ contain the maximum number of 0s and 1s. We will prove: for any two non-adjacent $r_{i}, r_{j}$ (non-adjacent means $i-j \neq \pm 1 (\bmod n)$), one of $r_{i}, r_{j}$ must be 0 or 1.

In fact, for any two non-adjacent $r_{i}, r_{j}$. Let $t=r_{i}+r_{j}$, and let $x=r_{i}$, then $\sum_{i=1}^{n} r_{i} r_{i+1}$ is a linear function of $x$. Therefore, when $\sum_{i=1}^{n} r_{i} r_{i+1}$ achieves its maximum value, it must be that $x \in \{0, \min \{1, t\}\}$. This indicates that one of $r_{i}, r_{j}$ must be 0 or 1.

The above conclusion shows: among $r_{1}, r_{2}, \ldots, r_{n}$, at most two numbers do not belong to $\{0,1\}$, and these two numbers must be adjacent.
If $r_{1}, r_{2}, \ldots, r_{n}$ are all 0 or 1, then there are exactly $r$ ones, so $\sum_{i=1}^{n} r_{i} r_{i+1} \leq r-1$,
If $r_{1}, r_{2}, \ldots, r_{n}$ have two numbers not in $\{0,1\}$, without loss of generality, let these be $r_{1}, r_{2} \in (0,1)$, then the remaining $n-2$ numbers are all 0 or 1. Thus, $r_{1} + r_{2} = 1$, and among the remaining $n-2$ numbers, there are exactly $r-1$ ones. Therefore, $\sum_{i=1}^{n} r_{i} r_{i+1} \leq r_{1} r_{2} + r_{2} + r - 2 + r_{1} = r - 3 + (1 + r_{1})(1 + r_{2}) \leq r - 3 + \frac{[(1 + r_{1}) + (1 + r_{2})]^2}{4} = r - \frac{3}{4}$.

In summary, the lemma is proved. And it is easy to see that the equality holds only when $r=n-1, r_{1}=r_{2}=\frac{1}{2}, r_{3}=r_{4}=\ldots=r_{n}=1$.
Now let's return to the original problem.
Let $\left[a_{i}\right]=b_{i}, \left\{a_{i}\right\}=r_{i} (1 \leq i \leq n)$, and let $\sum_{i=1}^{n} r_{i}=r$, then $r \in \mathbb{N}$, and $r \leq n-1$. According to the lemma, we have:
$$
\begin{array}{l}
\sum_{i=1}^{n}\left\{a_{i}\right\} a_{i+1}=\sum_{i=1}^{n} r_{i}\left(b_{i+1}+r_{i+1}\right)=\sum_{i=1}^{n} r_{i} b_{i+1}+\sum_{i=1}^{n} r_{i} r_{i+1} \\
\leq \sum_{i=1}^{n} b_{i+1}+\sum_{i=1}^{n} r_{i} r_{i+1} \leq \sum_{i=1}^{n} b_{i+1}+\sum_{i=1}^{n} r_{i}-\frac{3}{4}=n-\frac{3}{4}
\end{array}
$$

On the other hand, we take $a_{1}=\frac{3}{2}+(n-2) \varepsilon, a_{2}=\frac{1}{2}, a_{3}=a_{4}=\ldots=a_{n}=1-\varepsilon$. Then let $\varepsilon \rightarrow 0^{+}$,
then
$$
\sum_{i=1}^{n}\left\{a_{i}\right\} a_{i+1}=\left(\frac{1}{2}+(n-2) \varepsilon\right) \cdot \frac{1}{2}+\frac{1}{2} \cdot(1-\varepsilon)+(n-3)(1-\varepsilon)^{2}+(1-\varepsilon) \cdot\left(\frac{3}{2}+(n-\right.
$$
2) $\varepsilon) \rightarrow n-\frac{3}{4}$
In summary, the minimum value of $\lambda(n)$ is $n-\frac{3}{4}$.