Cronbach's alpha, variance vs. covariance, and how are negative values mathematically possible?

I'm trying to get my head around how a negative result from Cronbach's alpha is even mathematically possible. When I look at the formula, it's defined entirely in terms of variances of the individual items (the following from wikipedia): $$ {\displaystyle \alpha ={k \over k-1}\left(1-{\sum _{i=1}^{k}\sigma _{y_{i}}^{2} \over \sigma _{X}^{2}}\right)} $$ where: ${\displaystyle k}$ represents the number of "parts" (items, test parts, etc.) in the measure; the ${\displaystyle k/(k-1)}$ term causes alpha to be an unbiased estimate of reliability when the parts are parallel or essentially tau equivalent; ${\displaystyle \sigma _{y_{i}}^{2}}$ the variance associated with each part i; and ${\displaystyle \sigma _{X}^{2}}$ the observed score variance (the variance associated with the total test scores). It should not be possible for those variance terms to be negative. Variances can't be negative. They're derived from sums of squares, which are always positive, because you square the deviances before summing them. (mathematically, if the summed individual item variances were larger than the variance of the entire data set, then the ratio would be larger than 1, so subtracting that ratio from 1 would be negative. But is that even possible in a data set, that the summed variances of its subparts end up greater than the variance of the whole? That smells fishy to me, although I can't produce a proof that it's impossible.) However, all the questions I see here about negative Cronbach's a…