The 27th question: Find the smallest real number $\lambda$, such that $\left(a_{1}{ }^{n}+\prod_{i=1}^{n} a_{i}, a_{2}{ }^{n}+\prod_{i=1}^{n} a_{i}, \ldots, a_{n}{ }^{n}+\prod_{i=1}^{n} a_{i}\right) \leq$ $\lambda\left(a_{1}, a_{2}, \ldots, a_{n}\right)^{n}$ holds for any positive odd number $n$ and $n$ positive integers $a_{1}, a_{2}, \ldots, a_{n}$.
The 27th problem, solution: Without loss of generality, assume $\left(a_{1}, a_{2}, \ldots, a_{n}\right)=1$ (otherwise, replace $a_{i}$ with $\frac{a_{i}}{\left(a_{1}, a_{2}, \ldots, a_{n}\right)}$ for $1 \leq \mathrm{i} \leq \mathrm{n}$, which does not change the essence of the problem). In this case, we need to find the smallest real number $\lambda$ such that
$$
\begin{array}{l}
\left(a_{1}^{n}+\prod_{i=1}^{n} a_{i}, a_{2}^{n}+\prod_{i=1}^{n} a_{i}, \quad \ldots, a_{n}^{n}+\prod_{i=1}^{n} a_{i}\right) \leq \lambda \\
\text { Let }\left(a_{1}^{n}+\prod_{i=1}^{n} a_{i}, a_{2}^{n}+\prod_{i=1}^{n} a_{i}, \ldots, a_{n}^{n}+\prod_{i=1}^{n} a_{i}\right)=d, \text { then } \\
a_{1}^{n} \equiv a_{2}^{n} \equiv \ldots \equiv a_{n}^{n} \equiv-\prod_{i=1}^{n} a_{i}(\bmod d)
\end{array}
$$

We can prove that each $a_{i}(1 \leq \mathrm{i} \leq \mathrm{n})$ is coprime with $\mathrm{d}$.
If this conclusion is not true, assume without loss of generality that $\left(\mathrm{a}_{1}, \mathrm{~d}\right)>1$, and take any prime factor $\mathrm{p}$ of $\left(\mathrm{a}_{1}, \mathrm{~d}\right)$. Then, according to (2), we have:
$$
\begin{array}{l}
a_{1}^{n} \equiv a_{2}^{n} \equiv \ldots \equiv a_{n}^{n} \equiv-\prod_{i=1}^{n} a_{i}(\bmod p) \\
\Rightarrow a_{1} \equiv a_{2} \equiv \ldots \equiv a_{n} \equiv 0(\bmod p)
\end{array}
$$

This contradicts $\left(\mathrm{a}_{1}, \mathrm{a}_{2}, \ldots, \mathrm{a}_{\mathrm{n}}\right)=1$.
According to (2), we have:
$$
\prod_{i=1}^{n} a_{i}^{n} \equiv\left(-\prod_{i=1}^{n} a_{i}\right)^{n} \equiv-\prod_{i=1}^{n} a_{i}^{n}(\bmod d) \Rightarrow d \mid 2
$$

Thus, taking $\lambda=2$ in (1) satisfies the condition.
On the other hand, if $a_{1}=a_{2}=\ldots=a_{n}=1$, then $\lambda \geq 2$.
In conclusion, the smallest value of $\lambda$ that satisfies the condition is 2.