11. (20 points) Given
$$
f(x)=\frac{1+\ln (x+1)}{x}, g(x)=\frac{k}{x+1} .
$$

Find the largest positive integer $k$, such that for any positive number $c$, there exist real numbers $a$ and $b$ satisfying $-1<a<b<c$, and
$$
f(c)=f(a)=g(b) .
$$
11. For positive integer $k$, it is clear that $g(x)=\frac{k}{x+1}$ is a decreasing function on the interval $(-1,+\infty)$.
Thus, for any positive number $c$,
$$
f(c)=g(b)>g(c) \text {. }
$$

When $x>0$, the inequality
$$
\begin{array}{l}
f(x)>g(x) \\
\Leftrightarrow k0) \text {. }
$$

Then $h^{\prime}(x)=\frac{x-1-\ln (x+1)}{x^{2}}$.
$$
\text { Let } \varphi(x)=x-1-\ln (x+1)(x>0) \text {. }
$$

Then $\varphi^{\prime}(x)=\frac{x}{x+1}>0$.
Hence, $\varphi(x)$ is an increasing function when $x>0$.
Also, $\varphi(2)=1-\ln 30 \text {, }
$$

Therefore, there exists a unique positive real number $x_{0}$ such that
$$
\varphi\left(x_{0}\right)=x_{0}-1-\ln \left(x_{0}+1\right)=0 \text {. }
$$

Thus, $h^{\prime}\left(x_{0}\right)=0$, and $x_{0} \in(2,3)$.
Hence, when $x \in\left(0, x_{0}\right)$, $h^{\prime}(x)0$, $h(x)$ is an increasing function.

Therefore, when $x>0$, combining with equation (2), the minimum value of $h(x)$ is $h\left(x_{0}\right)=x_{0}+1 \in(3,4)$.
Combining with equation (1), we have the positive integer
$k \leqslant 3$.
We now prove that when $k=3$, for $-10 . \\
\text { Let } \tau(x)=1-2 x+(x+1) \ln (x+1) \text {, where } \\
\text { }-1\tau(0)>0$.
Therefore, equation (4) holds.
Note that,
$g(x)=\frac{3}{x+1}(x \in(-1,+\infty))$ has a range of $(0,+\infty)$,
$f(x)=\frac{1+\ln (x+1)}{x}(x \in(0,+\infty))$ also has a range of $(0,+\infty)$,
$$
f(x)=\frac{1+\ln (x+1)}{x}(x \in(-1,0)) \text { has a range }
$$

of $\mathbf{R}$.
Combining the graphs of the functions, we know that for any positive number $c$, there exist real numbers $a 、 b$ satisfying $-1<a<b<c$, and
$$
f(c)=f(a)=g(b) \text {. }
$$

In summary, the maximum value of the positive integer $k$ is 3.