The greatest $k$ is $13$.

Firstly, consider $a = b = \frac{2}{3}$ and $c = \frac{9}{4}$. They satisfy $abc = 1$. The inequality becomes
$$\frac{3}{2} + \frac{3}{2} + \frac{4}{9} + \frac{12k}{55} \ge \frac{k}{4} + 3,$$
i.e. $\frac{4}{9} \ge \frac{7k}{220}$. As $k$ is a positive integer, this implies $k \le 13$.

We now prove the inequality
$$\frac{1}{a} + \frac{1}{b} + \frac{1}{c} + \frac{13}{a+b+c+1} \ge \frac{13}{4} + 3 = \frac{25}{4}.$$
Let the left-hand side be $f(a, b, c)$. WLOG assume $c \ge b \ge a$. We first prove that
$$f(a, b, c) \ge f(\sqrt{ab}, \sqrt{ab}, c).$$
Note that the new triple still satisfies $(\sqrt{ab})(\sqrt{ab})(c) = 1$. Now,
$$\begin{align*} f(a, b, c) &\ge f(\sqrt{ab}, \sqrt{ab}, c) \\ \Leftrightarrow \quad & \frac{1}{a} + \frac{1}{b} - \frac{2}{\sqrt{ab}} \ge \frac{13}{2\sqrt{ab} + c + 1} - \frac{13}{a + b + c + 1} \\ \Leftrightarrow \quad & \frac{(\sqrt{a} - \sqrt{b})^2}{ab} \ge \frac{13(\sqrt{a} - \sqrt{b})^2}{(2\sqrt{ab} + c + 1)(a + b + c + 1)}. \end{align*}$$
It suffices to show $(2\sqrt{ab}+c+1)(a+b+c+1) \ge 13ab$. Indeed, since $a+b \ge 2\sqrt{ab}$, we only need to prove
$$2\sqrt{ab} + c + 1 \ge \sqrt{13ab}.$$
As $c \ge b \ge a$ and $abc = 1$, we have $c \ge 1$. Therefore, we have $\sqrt{c}(c+1) \ge 2$. This implies
$$c + 1 \ge \frac{2}{\sqrt{c}} = 2\sqrt{ab} > (\sqrt{13} - 2)\sqrt{ab}.$$
This proves $f(a, b, c) \ge f(\sqrt{ab}, \sqrt{ab}, c)$.

After this mixing step, it remains to consider the case $f\left(t, t, \frac{1}{t^2}\right)$ where $t > 0$. Now,
$$\begin{align*} f\left(t, t, \frac{1}{t^2}\right) &\ge \frac{25}{4} \\ \Leftrightarrow \quad & \frac{2}{t} + t^2 + \frac{13t^2}{2t^3 + t^2 + 1} \ge \frac{25}{4} \\ \Leftrightarrow \quad & 8t^6 + 4t^5 - 50t^4 + 47t^3 + 8t^2 - 25t + 8 \ge 0 \\ \Leftrightarrow \quad & (t-1)^2(8t^4 + 20t^3 - 18t^2 - 9t + 8) \ge 0. \end{align*}$$

It remains to prove $8t^4 + 20t^3 - 18t^2 - 9t + 8 \ge 0$ for $t > 0$. Indeed, the left-hand side is equal to
$$(8t^2 + 30t) \left(t - \frac{5}{8}\right)^2 + \frac{1}{32}(524t^2 - 663t + 256).$$
The first term is nonnegative since $t > 0$. Since $663^2 - 4(524)(256) < 0$, the last term is nonnegative. This proves the inequality and we are done.