21. 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 $M(\lambda)\left(\sum_{k=1}^{n} x_{k}\right)^{n} \geqslant \sum_{k=1}^{n} x_{k}^{n}+\lambda \prod_{k=1}^{n} x_{k}$.
21. 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 $M(\lambda) \geqslant 1$.
Thus, $M(\lambda) \geqslant \max \left\{1, \frac{1}{n^{n}}(n+\lambda)\right\}$.
We discuss two cases: (1) When $\frac{1}{n^{n}}(n+\lambda) \leqslant 1$, $\lambda \leqslant n^{n}-n$.
We need to prove $\left(\sum_{k=1}^{n} x_{k}\right)^{n} \geqslant \sum_{k=1}^{n} x_{k}^{n}+\left(n^{n}-n\right) \prod_{k=1}^{n} x_{k}$.
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 always $\left(x_{1} x_{2} \cdots x_{n}\right)^{n^{n} \cdots n}$, so,
(2) When $\frac{1}{n^{*}}(n+\lambda)>1$, $\lambda>n^{n}-n$.

We need to prove $\frac{n+\lambda}{n^{n}}\left(\sum_{k=1}^{n} x_{k}\right)^{n} \geqslant \sum_{k=1}^{n} x_{k}^{n}+\lambda \prod_{k=1}^{n} x_{k}$.
We only need to prove $\lambda\left[\left(\frac{1}{n} \sum_{k=1}^{n} x_{k}\right)^{n}-\prod_{k=1}^{n} x_{k}\right]+n\left(\frac{1}{n} \sum_{k=1}^{n} x_{k}\right)^{n} \geqslant \sum_{k=1}^{n} x_{k}^{n}$.
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 equation is equation (1).
In summary: The minimum value of $M(\lambda)$ is $\max \left\{1, \frac{1}{n^{n}}(n+\lambda)\right\}$.