Let $n$ be a natural number greater than 2. $l$ is a line on a plane. There are $n$ distinct points $P_1$, $P_2$, …, $P_n$ on $l$. Let the product of distances between $P_i$ and the other $n-1$ points be $d_i$ ($i = 1, 2,$ …, $n$). There exists a point $Q$, which does not lie on $l$, on the plane. Let the distance from $Q$ to $P_i$ be $C_i$ ($i = 1, 2,$ …, $n$). Find $S_n = \sum_{i = 1}^{n} (-1)^{n-i} \frac{c_i^2}{d_i}$.
1. **Assume the points are ordered on the line \( l \)**:
   Without loss of generality, we can assume that the points \( P_1, P_2, \ldots, P_n \) are arranged in this order on the line \( l \).

2. **Base case for \( n = 3 \)**:
   For \( n = 3 \), we can use Stewart's Theorem, which states that for a triangle \( \triangle ABC \) with a cevian \( AD \) (where \( D \) is on \( BC \)), the following relation holds:
   \[
   AB^2 \cdot DC + AC^2 \cdot BD = AD^2 \cdot BC + BD \cdot DC \cdot BC
   \]
   Applying this to our problem, we get:
   \[
   S_3 = (-1)^{3-1} \frac{c_1^2}{d_1} + (-1)^{3-2} \frac{c_2^2}{d_2} + (-1)^{3-3} \frac{c_3^2}{d_3} = 1
   \]
   Therefore, \( S_3 = 1 \).

3. **Case for \( n = 4 \)**:
   We need to compute \( S_4 \). Consider the points \( (P_1, P_2, P_4) \) and \( (P_1, P_3, P_4) \). We obtain the following system of equations:
   \[
   \left\{
   \begin{array}{ccc}
   \frac{c_1^2 \cdot P_1P_3}{d_1} - \frac{c_2^2 \cdot P_2P_3}{d_2} + \frac{c_4^2 \cdot P_4P_3}{d_4} = 1 \\
   \frac{c_1^2 \cdot P_1P_2}{d_1} - \frac{c_3^2 \cdot P_3P_2}{d_3} + \frac{c_4^2 \cdot P_4P_2}{d_4} = 1
   \end{array}
   \right.
   \]
   Subtracting the first equation from the second, we get:
   \[
   -P_2P_3 \cdot \frac{c_1^2}{d_1} + P_2P_3 \cdot \frac{c_2^2}{d_2} - P_2P_3 \cdot \frac{c_3^2}{d_3} + P_2P_3 \cdot \frac{c_4^2}{d_4} = 0
   \]
   Simplifying, we obtain:
   \[
   -\frac{c_1^2}{d_1} + \frac{c_2^2}{d_2} - \frac{c_3^2}{d_3} + \frac{c_4^2}{d_4} = 0
   \]
   Therefore, \( S_4 = 0 \).

4. **Induction hypothesis**:
   We claim that \( S_n = 0 \) for all \( n \geq 4 \). We will prove this by induction.

5. **Induction step**:
   Assume \( S_{n-1} = 0 \) for some \( n \geq 5 \). We need to show that \( S_n = 0 \).

   Consider the points \( (P_1, P_2, \ldots, P_{n-2}, P_n) \) and \( (P_1, P_2, \ldots, P_{n-3}, P_{n-1}, P_n) \). We obtain the following system of equations:
   \[
   \left\{
   \begin{array}{lcl}
   -P_{1}P_{n-1} \cdot \frac{c_1^2}{d_1} + P_{2}P_{n-1} \cdot \frac{c_2^2}{d_2} - \ldots + P_{n-3}P_{n-1} \cdot \frac{c_{n-3}^2}{d_{n-3}} - P_{n-2}P_{n-1} \cdot \frac{c_{n-2}^2}{d_{n-2}} + P_{n}P_{n-1} \cdot \frac{c_{n}^2}{d_{n}} = 0 \\
   -P_{1}P_{n-2} \cdot \frac{c_1^2}{d_1} + P_{2}P_{n-2} \cdot \frac{c_2^2}{d_2} - \ldots + P_{n-3}P_{n-2} \cdot \frac{c_{n-3}^2}{d_{n-3}} - P_{n-1}P_{n-2} \cdot \frac{c_{n-1}^2}{d_{n-1}} + P_{n}P_{n-2} \cdot \frac{c_{n}^2}{d_{n}} = 0
   \end{array}
   \right.
   \]
   Subtracting the first equation from the second, we get:
   \[
   P_{n-1}P_{n-2} \cdot \frac{c_1^2}{d_1} - P_{n-1}P_{n-2} \cdot \frac{c_2^2}{d_2} + \ldots - P_{n-1}P_{n-2} \cdot \frac{c_{n-3}^2}{d_{n-3}} + P_{n-1}P_{n-2} \cdot \frac{c_{n-2}^2}{d_{n-2}} - P_{n-1}P_{n-2} \cdot \frac{c_{n-1}^2}{d_{n-1}} + P_{n-1}P_{n-2} \cdot \frac{c_{n}^2}{d_{n}} = 0
   \]
   Simplifying, we obtain:
   \[
   S_n = 0
   \]
   Therefore, by induction, \( S_n = 0 \) for all \( n \geq 4 \).

\(\blacksquare\)

The final answer is \( \boxed{ S_n = 0 } \) for all \( n \geq 4 \).