Given a positive number $\lambda$ and a positive integer $n \geqslant 2$. Find the smallest positive number $M(\lambda)$, such that for all non-negative real numbers $x_{1}, x_{2} \cdots x_{n}$, we have $M(\lambda)\left(x_{1}+x_{2}+\cdots+\right.$ $\left.x_{n}\right)^{n} \geqslant x_{1}^{n}+x_{2}^{n}+\cdots+x_{n}^{n}+\lambda x_{1} x_{2} \cdots x_{n}$.
Let $x_{1}=x_{2}=\cdots=x_{n}=\frac{1}{n}$, from the inequality in the problem, we have $M(\lambda) \geqslant \frac{n}{n^{n}}+\frac{\lambda}{n^{n}}=\frac{1}{n^{n}}(n+\lambda)$.
Let $x_{1}=1, x_{2}=x_{3}=\cdots=x_{n}=0$, then we have $M(\lambda) \geqslant 1$.
Therefore, $M(\lambda) \geqslant \max \left\{1, \frac{1}{n^{n}}(n+\lambda)\right\}$.
We discuss two cases:
1. If $1 \geqslant \frac{n+\lambda}{n^{n}}, \lambda \leqslant n^{n}-n$
We need to prove $\left(x_{1}+x_{2}+\cdots+x_{n}\right)^{n} \geq x_{1}^{n}+x_{2}^{n}+\cdots+x_{n}^{n}+\left(n^{n}-n\right) x_{1} x_{2} \cdots x_{n}$ (1)
(If the above inequality holds, then $\geqslant x_{1}^{n}+x_{2}^{n}+\cdots+x_{n}^{n}+\lambda x_{1} x_{2} \cdots x_{n}$)
$\left(x_{1}+x_{2}+\cdots+x_{n}\right)^{n}-\left(x_{1}^{n}+x_{2}^{n}+\cdots+x_{n}^{n}\right)$ expands to $n^{n}-n$ terms.
Note: The product of these $n^{n}-n$ terms is certainly $\left(x_{1} x_{2} \cdots x_{n}\right)^{n^{n}-n}$.
So $\left(x_{1}+x_{2}+\cdots+x_{n}\right)^{n}-\left(x_{1}^{n}+x_{2}^{n}+\cdots+x_{n}^{n}\right)$
$\geqslant\left(n^{n}-n\right)^{n}-\sqrt[n]{\left(x_{1} x_{2} \cdots x_{n}\right)^{n_{n}}}=\left(n^{n}-n\right) x_{1} x_{2} \cdots x_{n}$.
2. If $1 < \frac{n+\lambda}{n^{n}}, \lambda > n^{n}-n$.
We need to prove $\frac{n+\lambda}{n^{n}}\left(x_{1}+x_{2}+\cdots+x_{n}\right)^{n} \geqslant x_{1}^{n}+x_{2}^{n}+\cdots+x_{n}^{n}+\lambda x_{1} x_{2} \cdots x_{n}$,

which is equivalent to $\lambda\left[\left(\frac{x_{1}+x_{2}+\cdots+x_{n}}{n}\right)^{n}-x_{1} x_{2} \cdots x_{n}\right]+n \cdot\left(\frac{x_{1}+x_{2}+\cdots+x_{n}}{n}\right)^{n}$
$\geqslant x_{1}^{n}+x_{2}^{n}+\cdots+x_{n}^{n}$
If we can prove:
$\left(n^{n}-n\right)\left[\left(\frac{x_{1} x_{2}+\cdots+x_{n}}{n}\right)^{n}-x_{1} x_{2} \cdots x_{n}\right]+n\left(\frac{x_{1}+x_{2}+\cdots+x_{n}}{n}\right)^{n}$
$\geqslant x_{1}^{n}+x_{2}^{n}+\cdots+x_{n}^{n}$, then the problem is solved.
The above inequality is $\left(x_{1}+x_{2}+\cdots+x_{n}\right)^{n} \geqslant x_{1}^{n}+x_{2}^{n}+\cdots+x_{n}^{n}+\left(n^{n}-n\right) x_{1} x_{2} \cdots x_{n}$, which completes the proof.
In conclusion, the minimum value of $M(\lambda)$ is $\max \left\{1, \frac{1}{n^{n}}(n+\lambda)\right\}$.