Bang, off with ya brain!

Take $(a_{j},b_{j})$ as points on complex plane. Then $Z_{j} = a_{j}+ib_{j}$ and $|Z_{j}|=\sqrt{a_{j}^{2}+b_{j}^{2}}$ . Now $|Z_{1}+Z_{2}| \leq |Z_{1}|+|Z_{2}|$ . This means $|\displaystyle \sum_{j=1}^{n}Z_{j}| \leq \displaystyle \sum_{j=1}^{n}|Z_{j}|$ . So minimum value of $\displaystyle \sum_{j=1}^{n}|Z_{j}|$ is $|\displaystyle \sum_{j=1}^{n}Z_{j}|$ . Now $\displaystyle \sum_{j=1}^{n}Z_{j}=\displaystyle \sum_{j=1}^{n}a_{j}+i\displaystyle \sum_{j=1}^{n}b_{j}=4-i$ . So minimum value of $(\displaystyle \sum_{j=1}^{n}\sqrt{a_{j}^{2}+b_{j}^{2}})^{2}$ is $|4-i|^{2}=17$

Note that this problem doesn't change for any value of $n$

View each pair $(a_i,b_i)$ as a ordered pair in the Cartesian plane.

Now each square root $\sqrt{a_i^2+b_i^2}$ represents the distance from the origin to that point (the hypotenuse).

Assume that we put each hypotenuse tip to tail so that they extend in a line. This "trail of hypotenuses" no matter what each individual length is, will end at the point $(p,q)$

Thus the overall length this trial will represent our sum. Which is obviously minimized when this length is straight. Or $\dfrac{a_i}{b_i}=\lambda$

So here the minimum value is $4^2+1^2=17$