Let $P=\{P_0,P_1,\ldots,P_8\}$ be the set of the $9$ points in the plane. Fix an arbitrary $P_0$. Now let $C$ be the number of triangles with vertices on the other $8$ points which include $P_0$ in their interior. We must show $C\equiv 0 \pmod{2}$. Now, fix some arbitrary $P_i$, $1\le i\le 8$. Now, draw line $P_0P_i$, and consider the pairs of the other $7$ points. To solve this problem, we will show that $C\equiv 0\pmod{2}$ remains invariant upon reflecting $P_i$ through $P_0$ (in other words, constructing $P_i'$ with the rule that the midpoint of $P_iP_i'$ is $P_0$). By doing so, we can reflect the right series of points to have all $8$ on the same half of the plane, and therefore since $C=0$ at this time, we have $C\equiv 0\pmod{2}$ for all formations of the $8$ points and arbitrary $P_0$. Now, if a pair of points $(P_x,P_y)$ lie on the same side of $P_0P_i$, then $\bigtriangleup P_iP_xP_y$ does not contain $P_0$ and neither does $\bigtriangleup P_i'P_xP_y$. So we can focus on only the pairs of points that have one on each side of line $P_0P_i$. Let $P_x$ and $P_y$ be on opposite sides of line $P_0P_i$. Now, notice that if $\bigtriangleup P_iP_xP_y$ contains $P_0$, then $\bigtriangleup P_i'P_xP_y$ does not. The converse of this statement is also true. Thus, let there be $a$ points on one side of line $P_0P_i$ and thus $7-a$ points on the other side of $P_0P_i$. Then let $B$ be the number of triangles with the above conditions that contain $P_0$. So, we have $a(7-a) - B$ triangles that do not contain $P_0$. But since $\displaystyle B\equiv a(7-a) - B\pmod{2}$ for all integral $a,B\ge 0$, then $C$'s parity is invariant. So we are done.

Assume that the set has $n$ points. Consider a point $P$ that is inside $k\geq1$ triangles whose vertices are in the set, and let $ABC$ be any such triangle. For each triangle $ABC$, and for each point in the set $Q\neq A,B,C,P$, either $A,B,C,Q$ are the vertices of a convex quadrilateral, and $P$ is inside exactly two of the four triangles formed by these vertices, one of which is $ABC$ (each one of the two diagonals of the quadrilateral divide it into two disjoint triangles, inside exactly one of both must be $P$), and if $A,B,C,Q$ are the vertices of a concave quadrilateral, then one of the triangles formed by these three vertices contains the fourth inside it, hence also $P$, while the other three possible triangles are disjoint and cover the previous triangle, hence exactly one of them has $P$ inside it, again exactly two of the four triangles formed by these vertices have $P$ inside it. Since this happens for each one of the $n-4$ points $Q$ that are not $A,B,C,P$, we conclude that, for each triangle $ABC$, there are exactly $n-4$ triangles that share exactly one side with $ABC$ and have $P$ inside it, plus $ABC$ itself, and no other triangle shares one side with $ABC$. Note that, if we count each edge $AB,BC,CA$ once for each triangle containing $P$ which admits it as a side, and add over the three of them, we obtain $n-4+3=n-1$. Adding over the $k$ triangles that have $P$ inside them, we obtain $k(n-1)$. Now, given any segment whose ends are points in the set other than $P$, let $u$ be the number of triangles that have this segment as one of its sides, and have $P$ inside them. For each one of these $u$ triangles, the segment is counted $u$ times in $k(n-1)$, hence $\sum u^2=k(n-1)$. But if we count the $3k$ edges of the $k$ triangles, we are counting each one of these segments $u$ times (as many times as it appears in a triangle), hence $\sum u=3k$. We conclude that \[(n-4)k=\sum u^2-\sum u=\sum u(u-1),\] where the sum is carried out over all possible segments formed by two points of the set other than $P$, and each term in the sum is clearly even, hence $(n-4)k$ is necessarily even. If $n$ is odd (e.g. $n=9$), then $k$ is even, and we are done. Note: One cannot say anything about the case when $n$ is even, since clearly $k=1$ would be odd if we consider a triangle with $P$ inside it for $n=4$, but in a regular $n$-gon with $n$ even, each point $P$ is inside $0$ (even) triangles.

Consider each of the $\binom{8}{3}$ triangles that may or may not contain $P$ as vertices of a graph. We draw a "graph edge" between two of these vertices (that is, triangles) if both triangles contain $P$ and they share a "geometrical edge" when considered as triangles and not vertices of the graph we are constructing (LOL). If a triangle $ABC$ contains $P$ on it's interior, draw the lines $AP,BP,CP$, and then convince yourself that this triangle will have 5 "graph edges" when considered as a vertex of the graph. If it does not contain $P$, then it will have 0 "graph edges". The number of vertices with odd degree on a graph is even. (I considered writing this without the edge distinction, for pure uninteligible awesomeness) Is there something wrong?

Of course, we're gonna change 9 for an arbitrary $n>1$ odd number. Let $A=\{P_1, P_2, \dots P_{n-1} \}$ be $n-1$ points on the plane. Our mission is to prove that, no matter where we put the last point $P_n$, it will be inside of an even number of triangles. We define the function $f: \mathcal{P} \backslash \{\overline{P_i P_j}, i, j \in \{ 1, 2, \dots n-1\} \} \rightarrow \mathbb{N}_0$ with $f(P) = \text{number of triangles with vertices in}\ A\ \text{which contain P}$. If we prove that $f$ is always even, we're done. Take the $n-1$ points and draw all the segments $\overline{P_iP_j}, i, j \in \{ 1, 2, \dots n-1\}$. They divide the plane in regions. If I have a region $\mathcal{R}$ and two points $A, B \in \mathcal{R}$, we have $f(A)=f(B)$, because we can move A to B without crossing any segment, so It wont cross any triangle boundary. Now, if we have two adyacent regions $\mathcal{R}, \mathcal{S}$, and two points $A \in \mathcal{R}, B \in \mathcal{S}$, the difference $f(A)-f(B)$ is even. Indeed, when we cross their common boundary, we´re changing the state (in-out) on a lot of triangles. In fact, we have $n-3$ triangles we're changing state, because we have two fixed points (the ones who delimite the boundary). So, if $r = \text{number of triangles which have common points with}\ \mathcal{R}$ and $s = \text{number of triangles which have common points with}\ \mathcal{S}$, it's pretty clear that $r+s=n-3$ and $f(A)-f(B)=r-s=r+s-2s=(n-3)-2s$ which is an even number. Now we're ready to prove the problem. If we start with a point $P$ very far (on the exterior region), it's clear that $f(P)=0$. Now, we can get to every region crossing some boundaries, and using the statement we prove, f(P) doesn't change it's parity, i.e, it's always even.