Colored graph with $2$ colors 
Let us call $G$ a graph with vertices in two possible colours. If we select a vertex, we change the colour of it and of every vertex that is adjacent to it. Is it possible to change a graph from all the first colour to all the second using such moves?

I cannot find any counter example but either cannot find a proof. What do you think? 
 A: The answer is YES and the above comments give several references for this result  called the "lamp lighting problem". Below there is a direct proof.
Let $G$ be a graph with vertices $v_1,v_2,\dots, v_n$ and let $A$ be a matrix such that $a_{ij}=1$ if and only if $i=j$ or there is an edge between $v_i$ and $v_j$. 
Assume that the vertices of $G$ are all blue (colour $0$). It is possible to change their colour to red (colour $1$) by using the those moves if and only if there is a vector $y=(y_1,\dots y_n)^t\in \mathbb{Z}_2^n$ such that
$$Ay=u$$
where $u=(1,1,\dots,1)^t$, that is, since the matrix $A$ is symmetric, if and only if 
$$u\in \text{Im}(A)=\text{Im}(A^t)=(\text{Ker}(A))^{\perp}.$$
Thus it suffices to show that $Ax=0$ implies $u\cdot x=0$:
$$\begin{align}
0&=\sum_{i=1}^n(Ax)_i=\sum_{i=1}^nx_i+\sum_{i=1}^n\sum_{j\not=i}a_{ij}x_j\\
&=u\cdot x+2\sum_{1\leq i<j\leq n}a_{ij}x_j= u\cdot x \pmod{2}.
\end{align}$$
and we are done.
A: Yes, it can be done for any graph. This is a much-discussed problem called the "lamp lighter's problem" or the "all ones problem" or "lights out". This paper which is available online looks like a good survey:
Rudolf Fleischer and Jiajin Yu, A Survey of the Game "Lights Out".
