An easy one for Romania TST!
We have:
(1) $ \sum_{k\in I\cup J}|u_{k}|^{2}= \sum_{k \in I \cap J} |u_k|^2 + \sum_{k \in I \setminus J} |u_k|^2 + \sum_{k \in J \setminus I} |u_k|^2 \geq $
$\geq \frac{1}{m}|\sum_{k \in I \cap J} u_k|^2 + \frac{1}{n-m}|\sum_{k \in I \setminus J} u_k|^2 + \frac{1}{n-m}|\sum_{k \in J \setminus I} u_k|^2$
Denote $x= |\sum_{k \in I \cap J} u_k| $. Then (1) can be continued as follows:
(2) $\cdots \geq \frac{x^2}{m} + \frac{2(1-x)^2}{n-m}$
RHS of (2) attains its min when $x_{\text{min}} =\frac{2m}{n+m}$ and an easy calculation shows that: $\frac{x_{\text{min}}^2}{m} + \frac{2(1-x_{\text{min}})^2}{n-m} = \frac{2}{n+m}$
So: $ \sum_{k\in I\cup J}|u_{k}|^{2} \geq \frac{2}{n+m} $
(1) and (2) become equalities when all vectors are collinear and with same directions and:
$u_k,\, k\in I \cap J$ are equal,
$u_k,\, k\in I-J $ are equal
$u_k,\, k\in J-I$ are equal.
$|\sum_{k \in I \cap J} u_k | = \frac{2m}{n+m}$
One can easily calculate the exact values of $u_k$