Is it possible to have two lines of best fit? Could you rig a data set to have two lines of equally good (and best) fit? Or is it impossible?
 A: It is certainly possible (in mostly silly cases).  For example, suppose you sample the same $x$ value twice and get two experimental $y$ values that differ.  One such case might be getting the points $(0, -1)$ and $(0, 1)$ if we sample $x=0$ twice.  Then, any line passing through the origin has SSE=2 which is the minimum possible.  But of course there are many lines passing through the origin.
The root source of this problem is the fact that the "best-fit-slope", the $\beta$ in the formula here: http://en.wikipedia.org/wiki/Simple_linear_regression#Fitting_the_regression_line is undefined (division by $0$).
A: The question is a bit unclear. If it is about uniqueness, this is a solution.
Consider the linear system
$$
\begin{align}
\mathbf{A} x &=b \\
%
\left(
\begin{array}{cc}
 1 & 0 \\
 0 & 0 \\
\end{array}
\right)
%
\left(
\begin{array}{cc}
 x \\
 y
\end{array}
\right)
%
&=
%
\left(
\begin{array}{l}
 b_{1} \\
 b_{2} 
\end{array}
\right)
\end{align}
$$
The least squares solution is defined as
$$
  x_{LS} = 
\left\{
  x \in \mathbb{R}^{2} \colon
\lVert 
  \mathbf{A}x - b
\rVert_{2}^{2}
\text{ is minimized}
\right\}
$$
Least squares minimizers form an affine space
The least squares minimizers for this problem are the affine space
$$
 x_{LS} = 
\color{blue}{\left(
\begin{array}{l}
 b_{1} \\
 0 
\end{array}
\right)}
+
\alpha
\color{red}{
\left(
\begin{array}{c}
 0 \\
 1 
\end{array}
\right)}, \qquad \alpha \in \mathbb{C}
$$
The coloring identifies $\color{blue}{range}$ and $\color{red}{null}$ space components.
Every point in this set generates the minimum value for the total error:
$$
  r^{2} \left( x_{LS} \right) = b_{1}^{*} b_{1}
$$
Solution of minimum norm
The norm of the minimizers is
$$
 \lVert x_{LS} \rVert_{2}^{2} = 
 \Bigg\lVert \color{blue}{\left(
\begin{array}{l}
 b_{1} \\
 0 
\end{array}
\right)}
%
  +
%
\alpha
\color{red}{
\left(
\begin{array}{c}
 0 \\
 1 
\end{array}
\right)}
 \Bigg\rVert_{2}^{2} = 
%
 \Bigg\lVert \color{blue}{\left(
\begin{array}{l}
 b_{1} \\
 0 
\end{array}
\right)}
 \Bigg\rVert_{2}^{2}
%
  +
%
\bar{\alpha}
 \Bigg\lVert_{2}^{2}
\color{red}{
\left(
\begin{array}{c}
 0 \\
 1 
\end{array}
\right)}
 \Bigg\rVert_{2}^{2} 
%
$$
The minimizer of minimum length, $\hat{x}$ is given by $\alpha = 0$, the projection onto $\color{blue}{\mathcal{R}\left( \mathbf{A}^{*} \right)}$. This is also the pseudoinverse solution
$$
 \color{blue}{\hat{x}} = 
 \color{blue}{\mathbf{A}^{\dagger} b}.
$$
