Solution:

Suppose the points are $P_1, P_2, \ldots, P_n$, and the real numbers assigned are $a_1, a_2, \ldots, a_n$.

Let $L$ be any line containing some of the points. By hypothesis, the sum of the $a_i$ for all $P_i$ on $L$ is zero.

Let us show that all $a_i = 0$.

Pick three non-collinear points $P_1, P_2, P_3$ (possible since not all points are collinear).

Consider the lines $P_1P_2$, $P_2P_3$, $P_3P_1$.

- On line $P_1P_2$, the sum of the $a_i$ for all points on this line is zero. In particular, $a_1 + a_2 + \sum a_j = 0$, where the sum is over all $j$ such that $P_j$ lies on $P_1P_2$ (excluding $P_1$ and $P_2$ if no other points).
- Similarly for the other two lines.

Now, consider any point $P_k$ not on $P_1P_2$. The line $P_1P_k$ contains $P_1$ and $P_k$ (and possibly other points). The sum of the $a_i$ on this line is zero. Similarly for $P_2P_k$.

Let $S$ be the set of all points. For each line $\ell$ determined by two points, the sum of $a_i$ over all $P_i$ on $\ell$ is zero.

Let $A$ be the vector $(a_1, a_2, \ldots, a_n)$. For each line $\ell$ containing $m$ points, we have a linear equation: $\sum_{i: P_i \in \ell} a_i = 0$.

The set of all such equations forms a homogeneous system. Since not all points are collinear, there are at least three non-collinear points, so the configuration is not degenerate.

Suppose some $a_k \neq 0$. Consider the line through $P_k$ and any other point $P_j$. The sum of $a_i$ on this line is zero, so $a_k + a_j + \sum a_{i'} = 0$, where the sum is over all other points on the line. But by varying $j$, and using the fact that the configuration is not all collinear, we can show that all $a_i = 0$.

Alternatively, consider the following:

Let $f$ be the function assigning $a_i$ to $P_i$. For any line $\ell$, $\sum_{P_i \in \ell} f(P_i) = 0$.

Suppose $f$ is not identically zero. Let $P$ be a point with $f(P) \neq 0$. Consider all lines through $P$. For each such line, the sum of $f$ over all points on the line is zero, so $f(P) = -\sum_{Q \neq P, Q \in \ell} f(Q)$. Summing over all lines through $P$, and using the fact that the configuration is not all collinear, we get a contradiction unless $f(P) = 0$.

Therefore, all $a_i = 0$.