4. Given $2 n$ positive real numbers $a_{1} \geqslant a_{2} \geqslant \cdots \geqslant a_{n}, b_{1}$ $\geqslant b_{2} \geqslant \cdots \geqslant b_{n}$ and $0 \leqslant \lambda \leqslant 2$, let
$$
\begin{array}{l}
M=\sum_{i=1}^{n} \sqrt{a_{i}^{2}+b_{i}^{2}-\lambda a_{i} b_{i}}, \\
N=\sum_{i=1}^{n} \sqrt{a_{i+1}^{2}+b_{i}^{2}-\lambda a_{i+1} b_{i}},
\end{array}
$$

where $a_{n+1}=a_{1}$.
Then the relationship between $M$ and $N$ is ().
(A) $M \geqslant N$
(B) $M \leqslant N$
(C) depends on $\lambda$
(D) completely uncertain
4. B.

Construct $\angle A O B$, with the included angle $\theta=\arccos \frac{\lambda}{2} \in[0, \pi]$.
As shown in Figure 1, take $n$ points $A_{1}$,
$A_{2}, \cdots, A_{n}$ and $B_{1}$,
$B_{2}, \cdots, B_{n}$ on the rays
$O A$ and $O B$ respectively, such that
$O A_{i}=a_{i}, O B_{i}=$
$b_{i}$. By the cosine rule,
we get $N \geqslant M$, that is,
$$
\begin{array}{l}
A_{2} B_{1}+A_{3} B_{2}+\cdots+A_{1} B_{n} \\
\geqslant A_{1} B_{1}+A_{2} B_{2}+\cdots+A_{n} B_{n} .
\end{array}
$$

Let $A_{1} B_{n}$ intersect $A_{2} B_{1}, A_{3} B_{2}, \cdots, A_{n} B_{n-1}$ at points $X_{1}, X_{2}, \cdots, X_{n-1}$ respectively. Then,
$$
\begin{array}{l} 
A_{2} B_{1}+A_{3} B_{2}+\cdots+A_{1} B_{n} \\
=\left(A_{1} X_{1}+B_{1} X_{1}\right)+\left(A_{2} X_{1}+X_{1} X_{2}+X_{2} B_{2}\right)+ \\
\cdots+\left(A_{n} X_{n-1}+X_{n-1} B_{n}\right) \\
\geqslant A_{1} B_{1}+A_{2} B_{2}+\cdots+A_{n} B_{n}
\end{array}
$$

we know the conclusion holds.