Extended $3$square problem We know about the famous 3-square problem. 

Draw three adjacent squares with same size and any two adjacent square has a common side. This looks like a rectangle. Name all of the vertices of square in anti clock wise from $A$ to $H$. Join $HB$, $HC$, and $HD$.
  Prove that
  $$\angle{ABH}+\angle{ACH}+\angle{ADH}=90^\circ$$  


Visual proof of 3-square problem. In this figure we can find three congruence right-angled triangles which are in yellowish-green color and their hands are in ratio $2:1$ and we can see two isosceles right angled triangle. Orange colored angle symbol denote half right angle (i.e,$45^\circ$).

In square $ABGH$, $HB$ is the diagonal of square. By using the property of diagonal of square we get, $\angle{ABH}=45^\circ$. We also know that $\angle{ABH}+\angle{ACH}+\angle{ADH}=90^\circ$.  
From above two equations, we get $\angle{ACH}+\angle{ADH}=45^\circ$. $\square$

Now, for my question:

Let $n$ be the number of squares. Is it possible, for $n$-squares, for the sum of all angles which lies below each line to became $180^\circ$?


 A: The question being asked is whether or not
$$S_n=\sum_{k=1}^n\arctan\left(1\over k\right)$$
is ever equal to $\pi$ (expressed in radians, or $180$, in degrees). By calculation one finds
$$S_{16}\approx3.1057646\lt\pi\lt3.16452042\approx S_{17}$$
so the answer is No.  It'd be nice, though, to have a less computationally-dependent answer (along the lines of the proof that $\sum_{k=1}^n{1\over k}$ is never an integer for $n\gt1$).
Remark (added later): The identity 
$$\arctan\left(1\over x\right)={\pi\over2}-\arctan x\quad\text{if }x\gt0$$
means you can relate this to a 2007 paper by Tewodros Amdeberhan, Luis A. Medina, and Victor H. Moll, in which they study the sequence
$$x_n=\tan\left(\sum_{k=1}^n\arctan k \right)$$
which, by virtue of the addition formula for the tangent function, satisfies the recursion equation
$$x_n={x_{n-1}+n\over1-nx_{n-1}}\quad\text{with }x_1=1$$
This gives the sequence
$$1,-3,0,4,-{9\over19},{105\over73},-{308\over331},{36\over43}\cdots,$$
The OEIS extends the sequences of numerators and denominators.
Amdeberhan et al. prove that $1-nx_{n-1}\not=0$ for $n\gt1$, so that the $x_n$'s are all well-defined rational numbers, and that $x_n\not=0$ for $n\ge4$. Together these imply that the OP's sum, $S_n$, is an integer multiple of $\pi\over2$ only for $n=3$, which is a way of ruling out $S_n=\pi$ without an explicit computation of $S_{16}$ and $S_{17}$. The proof, however, is not at all geometric.
