16. Let the function $f(x)=k x, g(x)=\frac{\ln x}{x}$. If the equation $f(x)=g(x)$ has two real solutions for $x$ in the interval $\left[\frac{1}{\mathrm{e}}, \mathrm{e}\right]$, then the range of the real number $k$ is
16. $k \in\left(\frac{1}{\mathrm{e}^{2}}, \frac{1}{2 \mathrm{e}}\right)$.

Notice that, $g(x)=\frac{\ln x}{x}$.
Then $g^{\prime}(x)=\frac{\frac{1}{x} \cdot x-\ln x}{x^{2}}=\frac{1-\ln x}{x^{2}}$.
It is easy to see that $g(x)$ is monotonically increasing on the interval $(0, \mathrm{e})$ and monotonically decreasing on the interval $(\mathrm{e},+\infty)$, achieving a minimum value at $x=\mathrm{e}$.

Thus, $k_{\text {min }}=\frac{\frac{1}{\mathrm{e}}}{\mathrm{e}}=\frac{1}{\mathrm{e}^{2}}$, and $g(x)$ is monotonically increasing on the interval $\left[\frac{1}{\mathrm{e}}, \mathrm{e}\right]$,
$$
g\left(\frac{1}{\mathrm{e}}\right)=-\mathrm{e}, g(\mathrm{e})=\frac{1}{\mathrm{e}}, g(1)=0 .
$$

When $f(x)$ and $g(x)$ have only one intersection point on the interval $\left[\frac{1}{\mathrm{e}}, \mathrm{e}\right]$, i.e., the graph of $f(x)$ is tangent to the graph of $g(x)$, $k$ takes its maximum value.
Assume the coordinates of the tangent point are $\left(x_{0}, k x_{0}\right)$, the slope is
$$
k=\frac{1-\ln x_{0}}{x_{0}^{2}} \text {. }
$$

Since the point $\left(x_{0}, k x_{0}\right)$ is on $g(x)$, we have,
$$
k x_{0}=\frac{\ln x_{0}}{x_{0}} .
$$

Solving equations (1) and (2) together, we get $x_{0}=\sqrt{\mathrm{e}}, k=\frac{1}{2 \mathrm{e}}$. Therefore, $k \in\left[\frac{1}{\mathrm{e}^{2}}, \frac{1}{2 \mathrm{e}}\right)$.