4. As shown in Figure $3, M$ and $N$ are internal points on the diagonals $A C$ and $C E$ of the regular hexagon $A B C D E F$, and $\frac{A M}{A C}=\frac{C N}{C E}=\lambda$. If points $B, M, N$ are collinear, then $\lambda$ $=$
4. $\frac{\sqrt{3}}{3}$.

Extend $E A$ and $C B$ to intersect at point $P$, and let the side length of the regular hexagon be 1.
It is easy to know that $P B=2, A$ is the midpoint of $E P$, $E A=A P=\sqrt{3}$.
From $A M=\lambda A C$, we get
$$
C M=(1-\lambda) C A \text {. }
$$

Also, $C P=3 C B, C A$ is the median of $\triangle P C E$ on side $P E$, $C N$ $=\lambda \boldsymbol{C E}$, then we have
$$
C A=\frac{1}{2}(C E+C P),
$$

which means $\frac{1}{1-\lambda} C M=\frac{1}{2 \lambda} C N+\frac{3}{2} C B$.
Rearranging gives $C M=\frac{1-\lambda}{2 \lambda} C N+\frac{3(1-\lambda)}{2} C B$.
Since when $B, M, N$ are collinear, there exists a real number $t$ such that
$C M=(1-t) C N+t C B$,
thus $\frac{1-\lambda}{2 \lambda}+\frac{3(1-\lambda)}{2}=1$.
Solving for $\lambda$ gives $\lambda=\frac{\sqrt{3}}{3}$.