Find the smallest constant $\lambda$ such that
$$
a_{1} a_{2} \cdots a_{n}\left(\frac{1}{a_{1}}+\frac{1}{a_{2}}+\cdots+\frac{1}{a_{n}}-n+\lambda\right) \leqslant \lambda \text {, }
$$

for any integer $n \geq 2$ and any sequence $\left(a_{1}, a_{2}, \cdots, a_{n}\right)$ satisfying
$$
\begin{array}{l}
a_{1}, a_{2}, \cdots, a_{n} \in \mathbf{R}_{+}, \\
a_{1}+a_{2}+\cdots+a_{n}=n
\end{array}
$$
Solve for the required $\lambda_{\text {min }}=e$.
Define
$$
\begin{aligned}
E_{n}\left(a_{1}, a_{2}, \cdots, a_{n}\right) \\
=\frac{\lambda}{a_{1} a_{2} \cdots a_{n}}-\lambda+\left(n-\frac{1}{a_{1}}-\frac{1}{a_{2}}-\cdots-\frac{1}{a_{n}}\right) .
\end{aligned}
$$

First, prove: $\lambda \geqslant e$.
Notice that the original inequality holds if and only if $E_{n} \geqslant 0$.
Take $a_{1}=a_{2}=\cdots=a_{n-1}=t\left(1 \leqslant t0\right)$.
When $t=1$, the original inequality clearly holds, so we may assume $t>1$. In this case, $E_{n}(t, t, \cdots, t, n-(n-1) t) \geqslant 0$, simplifying the calculation, we get
the original inequality
$$
\begin{aligned}
&\Leftrightarrow \lambda \geqslant \frac{n(n-1) t^{n-2}(t-1)^{2}}{(n-1) t^{n}-n t^{n-1}+1} . \\
&\text { Let } f(t)=\frac{n(n-1) t^{n-2}(t-1)^{2}}{(n-1) t^{n}-n t^{n-1}+1} \\
&=\frac{n(n-1)\left(1-\frac{1}{t}\right)^{2}}{(n-1)-\frac{n}{t}+\frac{1}{t^{n}}} .
\end{aligned}
$$

It can be verified that $f(t)$ is monotonically increasing, and since $t$ is bounded, $f(t)$ is also bounded.
$$
\begin{aligned}
&\text { Then } \lambda \geqslant \lim _{1 \rightarrow \frac{n}{n-1}} f(t)=\left(1+\frac{1}{n-1}\right)^{n-1} \\
&\Rightarrow \lambda \geqslant \lim _{n \rightarrow+\infty}\left(1+\frac{1}{n-1}\right)^{n-1}=e .
\end{aligned}
$$

Next, prove: When $\lambda=e$, the original inequality holds.
The proof uses a combination of the adjustment method and mathematical induction.
When $n=2$, we need to prove
$$
\begin{aligned}
& a_{1} a_{2}\left(\frac{1}{a_{1}}+\frac{1}{a_{2}}-2+e\right) \leqslant e\left(a_{1}+a_{2}=2\right) . \\
& \text { And } a_{1} a_{2}\left(\frac{1}{a_{1}}+\frac{1}{a_{2}}-2+e\right) \leqslant e \\
&\Leftrightarrow 2\left(a_{1}-1\right)^{2} \leqslant e\left(a_{1}-1\right)^{2},
\end{aligned}
$$

which is clearly true.
Assume that for $n=k(k \geqslant 2)$, the proposition holds, i.e., for
$$
\begin{aligned}
& a_{1}, a_{2}, \cdots, a_{k} \in \mathbf{R}_{+}, \\
& a_{1}+a_{2}+\cdots+a_{k}=k,
\end{aligned}
$$

we have $E_{k}\left(a_{1}, a_{2}, \cdots, a_{k}\right) \geqslant 0$.
For $a_{1}, a_{2}, \cdots, a_{k+1} \in \mathbf{R}_{+}$,
$$
a_{1}+a_{2}+\cdots a_{k+1}=k+1,
$$

by symmetry, we can assume
$$
a_{1}=\max _{1 \leq i0.
$$
Thus, inequality (2) $\Leftrightarrow \lambda \geqslant\left(a_{1}+a_{k+1}\right) a_{2} a_{3} \cdots a_{k}$
By the AM-GM inequality,
$\left(a_{1}+a_{k+1}\right) a_{2} a_{3} \cdots a_{k}$
$$
\leqslant\left(\frac{\sum_{i=1}^{k+1} a_{i}}{k}\right)^{k}=\left(\frac{k+1}{k}\right)^{k}=\left(1+\frac{1}{k}\right)^{k}<e=\lambda.
$$

Therefore, $E_{k+1}\left(a_{1}, a_{2}, \cdots, a_{k+1}\right)$
$$
\begin{aligned}
& \geqslant E_{k+1}\left(a_{1}+a_{k+1}-1, a_{2}, \cdots, a_{k}, 1\right) \\
& =E_{k}\left(a_{1}+a_{k+1}-1, a_{2}, \cdots, a_{k}\right) .
\end{aligned}
$$

By the induction hypothesis, $E_{k} \geqslant 0$, so when $n=k+1$,
$$
E_{k+1}\left(a_{1}, a_{2}, \cdots, a_{k+1}\right) \geqslant 0.
$$

Therefore, when $n=k+1$, the proposition also holds.
By induction, when $\lambda=e$, the original inequality holds.