In some degenerate cases there may be no such a one point (for instance, if all the lines are parallel). However there's a single solution in the general case.
I assume you're trying to solve a more general problem where all the lines are not required to intersect exactly (otherwise there's a much simpler solution than the least squares).
Derivation:
You say the every line is represented by two points. Let's rather work in the convention where a line is represented by one point and a direction vector, which is just a vector subtraction of those two points. That is, instead of describing a line by points $\mathbf{a}$ and $\mathbf{b}$ we'll describe it by a point $\mathbf{a}$ and a vector $\mathbf{d}$ whereas $\mathbf{d}=\mathbf{b}-\mathbf{a}$.
Our point (which we're trying to find) is $\mathbf{c}$.
The distance of this point to the line is:
$H=\frac{\|(\mathbf{c}-\mathbf{a})\times\mathbf{d}\|}{\|\mathbf{d}\|}$
Using identity $(\mathbf{a}\times\mathbf{b})\cdot(\mathbf{a}\times\mathbf{b})=\|\mathbf{a}\|^2\|\mathbf{b}\|^2-(\mathbf{a}\cdot\mathbf{b})^2$
we have:
$H^2=\frac{\|\mathbf{c}-\mathbf{a}\|^2\|\mathbf{d}\|^2-\|(\mathbf{c}-\mathbf{a})\cdot\mathbf{d}\|^2 }{\|\mathbf{d}\|^2}$
$H^2 = \|\mathbf{c}-\mathbf{a}\|^2-\frac{\|(\mathbf{c}-\mathbf{a})\cdot\mathbf{d}\|^2 }{\|\mathbf{d}\|^2}$
The square sum of the distances of the point $\mathbf{c}$ to all the lines is just the sum of the above expressions for all the lines. The problem is to minimize this sum. This sum depends on a variable $\mathbf{c}$ (which is actually 3 variables, the components of $\mathbf{c}$). This is a standard least squares problem, which generally has a single solution (unless there's a degeneracy).
Solving the least squares for this specific case.
Since we want find such a $\mathbf{c}$ that minimizes this sum, its derivative with regard to $\mathbf{c}$ should be zero. In other words:
$\frac{d(H^2)}{d\mathbf{c}}=2(\mathbf{c}-\mathbf{a})-2\mathbf{d}\frac{(\mathbf{c}-\mathbf{a})\cdot\mathbf{d}}{\|\mathbf{d}\|^2}$
$0=\sum_{i=0}^m{\mathbf{c}-\mathbf{a}^{(i)}-\mathbf{d}^{(i)}\frac{(\mathbf{c}-\mathbf{a}^{(i)})\cdot\mathbf{d}^{(i)}}{\|\mathbf{d}^{(i)}\|^2}}$
This gives 3 equations (since it's a vector equation) with 3 unknowns (components of $\mathbf{c}$).
Best Answer
In order to get to the minimum distance problem quickly, we quote a formula for the distance from the point $(x_0,y_0)$ to the line with equation $px+qy+r=0$. This distance is $$\frac{|px_0+qy_0+r|}{\sqrt{p^2+q^2}}.$$ The site linked to gives a proof, as does Wikipedia. The distance from a point to a line has also been discussed more than once on StackExchange.
For our minimization problem, we can assume that one of the points is $(0,0)$, and the other is, say, $(a,b)$. So the line has equation $bx-ay=0$. We can assume that $a$ and $b$ are relatively prime, for if $d$ is their greatest common divisor, the same line is determined by $(0,0)$ and $(a/d,b/d)$.
Then by Bezout's Theorem, we can find integers $x_0$, $y_0$ between $0$ and $\max(a,b)$ such that $|bx_0-ay_0|=1$. Thus the smallest non-zero value of $|bx-ay|$ as $(x,y)$ ranges over our grid is $1$.
So the minimum possible non-zero distance from a gridpoint to our line is $\dfrac{1}{\sqrt{a^2+b^2}}$.
Now we want to maximize $a^2+b^2$, as $(a,b)$ ranges over relatively prime pairs on our grid. Then for fixed $a+b$, this maximum occurs when $a$ and $b$ differ by as little as possible. If $n=1$, the maximum value of $\sqrt{a^2+b^2}$ is achieved when $a=b=1$.
Suppose now that $n>1$. Then the maximum value of $a^2+b^2$, given that $\gcd(a,b)=1$ and $(a,b)$ is on our grid is attained at $a=n-1$, $b=n$, and at $a=n$, $b=n-1$. The minimum non-zero distance is therefore $$\frac{1}{\sqrt{2n^2-2n+1}}.$$
Comment: For our particular problem, we do not need Bezout's Theorem, since if $n>1$ then the distance from $(1,1)$ to the line through $(0,0)$ and $(n-1,n)$ is $1/(2n^2-2n+1)$. It is clear that we can't do better, since $|bx_0-ay_0|$ is at least $1$. However, if the $n\times n$ grid is replaced by an $m\times n$ grid, the natural analysis uses Bezout's Theorem.