Why Spearman's rank correlation ranges from from -1 to 1 $$\rho = 1 - \frac{6 \sum d_i^2}{n(n^2 - 1)}$$
$\rho$ = Spearman's rank correlation coefficient
$d_i$ = difference between the two ranks of each observation
$n$ = number of observations
Given the Spearman's rank correlation above, it's clear to see the maximum is 1 as the smallest $d_i$ will be zero. I am trying to figure out why it is -1 by having two ranks in exactly reversed order.
I attach my calculation below, where the red rectangle represents the specific case of four rows, and I ordered two ranks in exactly reverse order.
The purple rectangle is the actual calculation, and it indeed results in -1.
I try to generalize in the green rectangle, but how does that formula result in 2?


As you note the sum includes $1^2+3^2+5^2+\cdots+(m-1)^2$ or $\sum_1^{m/2} (2r-1)^2$. For simplicity I'll look at the case when there are an even number of rows, ie when $m$ is even.
You can expand the sum by expanding bracket as $$\sum_1^{m/2} (2r-1)^2 = 4\sum r^2-4\sum r +\sum 1$$ and use standard formulae for the sums: $\sum_1^n r^2 = \frac16 n(n+1)(2n+1)$ and $\sum_1^n r =\frac12 n(n+1)$
This gives $$\frac46 (m/2)((m/2)+1)(m+1) - \frac42 m/2(m/2+1) + m/2= \frac16 (m^3-m).$$
We can now substitute this into your expression:
$$\frac{6\cdot(2\sum_1^{m/2} (2r-1)^2)}{m^3-m}$$
$$=\frac{6\cdot(2\cdot \frac16 (m^3-m)}{m^3-m}$$
$$=\frac{2 (m^3 -m)}{m^3-m}$$
$$=2$$
And so the Spearman rank coefficient is -1
So in this case, one can directly calculate the value.  The standard formulae may not be familiar to you. Typically they are proved by induction, but a visual proof is offered on maths StackExchange: https://math.stackexchange.com/questions/122546/gaussian-proof-for-the-sum-of-squares
