Denote the coordinates of $M$ and $N$ as $(x_1, y_1)$ and $(x_2, y_2)$ respectively. Denote the tangent lines of $C$ at $M$ and $N$ by $l_1$ and $l_2$ respectively, with their intersection point being $P(x_p, y_p)$. Suppose the slope ratio of line $l$ is $k$. Then we can write the equation of $l$ as $y = kx + 1$.

Eliminating $y$ from
$$\begin{cases} y = x + \frac{1}{x}, \\ y = kx + 1, \end{cases}$$
we get $x + \frac{1}{x} = kx + 1$, i.e. $(k-1)x^2 + x - 1 = 0$. By the assumption, we know that the equation has two distinctive real roots, $x_1$ and $x_2$, on $(0, +\infty)$. Then $k \neq 1$, and
$$\Delta = 1 + 4(k - 1) > 0,$$
$$x_1 + x_2 = \frac{1}{1-k} > 0,$$
$$x_1 x_2 = \frac{1}{1-k} > 0.$$
From the above we get $\frac{3}{4} < k < 1$.

We find the derivative of $y = x + \frac{1}{x}$ as $y' = 1 - \frac{1}{x^2}$. Then $y'|_{x=x_1} = 1 - \frac{1}{x_1^2}$ and $y'|_{x=x_2} = 1 - \frac{1}{x_2^2}$. Therefore, the equation of line $l_1$ is
$$y - y_1 = \left(1 - \frac{1}{x_1^2}\right)(x - x_1)$$
or
$$y - \left(x_1 + \frac{1}{x_1}\right) = \left(1 - \frac{1}{x_1^2}\right)(x - x_1).$$
After simplification, we get
$$y = \left(1 - \frac{1}{x_1^2}\right)x + \frac{2}{x_1}$$
In the same way, we get the equation of $l_2$,
$$y = \left(1 - \frac{1}{x_2^2}\right)x + \frac{2}{x_2}$$
By subtracting, we get
$$\left(\frac{1}{x_2^2} - \frac{1}{x_1^2}\right)x_p + \frac{2}{x_1} - \frac{2}{x_2} = 0.$$
Since $x_1 \neq x_2$, we have
$$x_p = \frac{2x_1x_2}{x_1 + x_2}$$
Substituting $x_1 + x_2$ and $x_1 x_2$ from above, we obtain $x_p = 2$.

By adding, we obtain
$$2y_p = \left(2 - \left(\frac{1}{x_1^2} + \frac{1}{x_2^2}\right)\right)x_p + 2\left(\frac{1}{x_1} + \frac{1}{x_2}\right)$$
where
$$\frac{1}{x_1} + \frac{1}{x_2} = \frac{x_1 + x_2}{x_1 x_2} = 1,$$
$$\begin{aligned} \frac{1}{x_1^2} + \frac{1}{x_2^2} &= \frac{x_1^2 + x_2^2}{x_1^2 x_2^2} = \frac{(x_1 + x_2)^2 - 2x_1x_2}{x_1^2 x_2^2} \\ &= \left(\frac{x_1 + x_2}{x_1 x_2}\right)^2 - \frac{2}{x_1 x_2} = 2k - 1. \end{aligned}$$
Substituting it into the previous equation, we have $2y_p = (3 - 2k)x_p + 2$. Since $x_p = 2$, then $y_p = 4 - 2k$. As $\frac{3}{4} < k < 1$, we get $2 < y_p < \frac{5}{2}$.

Therefore, the locus of point $P$ is the segment between $(2, 2)$ and $(2, 2.5)$ (not including the endpoints).