The length of the diagonal is $\sqrt{3^2 + 4^2} = 5$ (a 3-4-5 [right triangle](https://artofproblemsolving.com/wiki/index.php/Right_triangle)). For each $k$, $\overline{P_kQ_k}$ is the [hypotenuse](https://artofproblemsolving.com/wiki/index.php/Hypotenuse) of a $3-4-5$ right triangle with sides of $3 \cdot \frac{168-k}{168}, 4 \cdot \frac{168-k}{168}$. Thus, its length is $5 \cdot \frac{168-k}{168}$. Let $a_k=\frac{5(168-k)}{168}$. We want to find $2\sum\limits_{k=1}^{168} a_k-5$ since we are over counting the diagonal.
$2\sum\limits_{k=1}^{168} \frac{5(168-k)}{168}-5 =2\frac{(0+5)\cdot169}{2}-5 =168\cdot5 =\boxed{840}$
[Rectangle](https://artofproblemsolving.com/wiki/index.php/Rectangle) $ABCD_{}^{}$ has [ sides](https://artofproblemsolving.com/wiki/index.php/Edge) $\overline {AB}$ of [length](https://artofproblemsolving.com/wiki/index.php/Length) 4 and $\overline {CB}$ of length 3. Divide $\overline {AB}$ into 168 [congruent](https://artofproblemsolving.com/wiki/index.php/Congruent) [segments](https://artofproblemsolving.com/wiki/index.php/Segment) with [points](https://artofproblemsolving.com/wiki/index.php/Point) $A_{}^{}=P_0, P_1, \ldots, P_{168}=B$, and divide $\overline {CB}$ into 168 congruent segments with points $C_{}^{}=Q_0, Q_1, \ldots, Q_{168}=B$. For $1_{}^{} \le k \le 167$, draw the segments $\overline {P_kQ_k}$. Repeat this [construction](https://artofproblemsolving.com/wiki/index.php/Construction) on the sides $\overline {AD}$ and $\overline {CD}$, and then draw the [diagonal](https://artofproblemsolving.com/wiki/index.php/Diagonal) $\overline {AC}$. Find the sum of the lengths of the 335 [parallel](https://artofproblemsolving.com/wiki/index.php/Parallel) segments drawn.