5. For the ellipse $\frac{x^{2}}{9}+\frac{y^{2}}{4}=1$, the right focus is $F$, and the right directrix is $l$. Points $P_{1}, P_{2}, \cdots, P_{24}$ are 24 points arranged in a counterclockwise order on the ellipse, where $P_{1}$ is the right vertex of the ellipse, and
$$
\angle P_{1} F P_{2}=\angle P_{2} F P_{3}=\cdots=\angle P_{23} F P_{24}=\angle P_{24} F P_{1} .
$$

Then the sum of the reciprocals of the distances from these 24 points to $l$ is
$5.6 \sqrt{5}$.
In the ellipse, $F(\sqrt{5}, 0), e=\frac{\sqrt{5}}{3}, l: x=\frac{9}{\sqrt{5}}$.
It is easy to know that for each $i(i=1,2, \cdots, 12)$, points $P_{i} 、 F 、 P_{i+12}$ are collinear.
Let $l_{P_{i} P_{i+12}}: y=k_{i}(x-\sqrt{5})$, substituting into $\frac{x^{2}}{9}+\frac{y^{2}}{4}=1$, we get $\left(4+9 k_{i}^{2}\right) x^{2}-18 \sqrt{5} k_{i}^{2} x+45 k_{i}^{2}-36=0$.
Then $x_{i}+x_{i+12}=\frac{18 \sqrt{5} k_{i}^{2}}{4+9 k_{i}^{2}}$,
$$
x_{i} x_{i+12}=\frac{45 k_{i}^{2}-36}{4+9 k_{i}{ }^{2}}(i=1,2, \cdots, 12) \text {. }
$$

The distances from points $P_{i} 、 P_{i+12}$ to $l$ are respectively
$$
d_{i}=\frac{9}{\sqrt{5}}-x_{i}, d_{i+12}=\frac{9}{\sqrt{5}}-x_{i+12} \text {, }
$$

Then $\frac{1}{d_{i}}+\frac{1}{d_{i+12}}=\frac{1}{\frac{9}{\sqrt{5}}-x_{i}}+\frac{1}{\frac{9}{\sqrt{5}}-x_{i+12}}$
$$
=\frac{\frac{18}{\sqrt{5}}-\left(x_{i}+x_{i+12}\right)}{\frac{81}{5}-\frac{9}{\sqrt{5}}\left(x_{i}+x_{i+12}\right)+x_{i} x_{i+12}} \text {. }
$$

Substituting the expressions for $x_{i}+x_{i+12}$ and $x_{i} x_{i+12}$ into the above equation, we get
$$
\frac{1}{d_{i}}+\frac{1}{d_{i+12}}=\frac{\sqrt{5}}{2} \text {. }
$$

Therefore, the sum of the reciprocals of the distances from these 24 points to $l$ is
$$
\sum_{i=1}^{12}\left(\frac{1}{d_{i}}+\frac{1}{d_{i+12}}\right)=12 \times \frac{\sqrt{5}}{2}=6 \sqrt{5} .
$$