Given a positive number $\lambda$ and a positive integer $n(n \geqslant$ 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
$$
M(\lambda)\left(\sum_{k=1}^{n} x_{k}\right)^{n} \geqslant \sum_{k=1}^{n} x_{k}^{n}+\lambda \prod_{k=1}^{n} x_{k} .
$$
Let $x_{1}=x_{2}=\cdots=x_{n}=\frac{1}{n}$, then 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) When $\frac{1}{n^{n}}(n+\lambda) \leqslant 1$, $\lambda \leqslant n^{n}-n$.

We need to prove
$$
\left(\sum_{k=1}^{n} x_{k}\right)^{n} \geqslant \sum_{k=1}^{n} x_{k}^{n}+\left(n^{n}-n\right) \prod_{k=1}^{n} x_{k} \text {. }
$$

Since $\left(\sum_{k=1}^{n} x_{k}\right)^{n}-\sum_{k=1}^{n} x_{k}^{n}$ expands to $n^{n}-n$ terms, and

the product of these $n^{n}-n$ terms is certainly $\left(x_{1} x_{2} \cdots x_{n}\right)^{n^{n}-n}$, so,
$$
\begin{array}{l}
\left(\sum_{k=1}^{n} x_{k}\right)^{n}-\sum_{k=1}^{n} x_{k}^{n} \\
\geqslant\left(n^{n}-n\right) \cdot \sqrt[n^{n}-n]{\left(x_{1} x_{2} \cdots x_{n}\right)^{n^{n}-n}} \\
=\left(n^{n}-n\right) \prod_{k=1}^{n} x_{k} \geqslant \lambda \prod_{k=1}^{n} x_{k} .
\end{array}
$$
(2) When $\frac{1}{n^{n}}(n+\lambda)>1$, $\lambda>n^{n}-n$.

We need to prove $\frac{n+\lambda}{n^{n}}\left(\sum_{k=1}^{n} x_{k}\right)^{n} \geqslant \sum_{k=1}^{n} x_{k}^{n}+\lambda \prod_{k=1}^{n} x_{k}$,
which is equivalent to proving
$$
\lambda\left[\left(\frac{1}{n} \sum_{k=1}^{n} x_{k}\right)^{n}-\prod_{k=1}^{n} x_{k}\right]+n\left(\frac{1}{n} \sum_{k=1}^{n} x_{k}\right)^{n} \geqslant \sum_{k=1}^{n} x_{k}^{n} \text {. }
$$

If we can prove
$$
\left(n^{n}-n\right)\left[\left(\frac{1}{n} \sum_{k=1}^{n} x_{k}\right)^{n}-\prod_{k=1}^{n} x_{k}\right]+n\left(\frac{1}{n} \sum_{k=1}^{n} x_{k}\right)^{n} \geqslant \sum_{k=1}^{n} x_{k}^{n},
$$

then the problem is solved. The above inequality is the same as inequality (1).
In summary, the minimum value of $M(\lambda)$ is
$$
\max \left\{1, \frac{1}{n^{n}}(n+\lambda)\right\} \text {. }
$$