Let $m<n$ be two positive integers, let $I$ and $J$ be two index sets such that $|I|=|J|=n$ and $|I\cap J|=m$, and let $u_k$, $k\in I\cup J$ be a collection of vectors in the Euclidean plane such that \[|\sum_{i\in I}u_i|=1=|\sum_{j\in J}u_j|.\] Prove that \[\sum_{k\in I\cup J}|u_k|^2\geq \frac{2}{m+n}\] and find the cases of equality.
1. We start by considering the given conditions:
   \[
   \left|\sum_{i \in I} u_i\right| = 1 \quad \text{and} \quad \left|\sum_{j \in J} u_j\right| = 1
   \]
   where \( |I| = |J| = n \) and \( |I \cap J| = m \).

2. We need to prove that:
   \[
   \sum_{k \in I \cup J} |u_k|^2 \geq \frac{2}{m+n}
   \]

3. We decompose the sum of the squared magnitudes of the vectors:
   \[
   \sum_{k \in I \cup J} |u_k|^2 = \sum_{k \in I \cap J} |u_k|^2 + \sum_{k \in I \setminus J} |u_k|^2 + \sum_{k \in J \setminus I} |u_k|^2
   \]

4. By the Cauchy-Schwarz inequality, we have:
   \[
   \left|\sum_{k \in I \cap J} u_k\right|^2 \leq m \sum_{k \in I \cap J} |u_k|^2
   \]
   \[
   \left|\sum_{k \in I \setminus J} u_k\right|^2 \leq (n-m) \sum_{k \in I \setminus J} |u_k|^2
   \]
   \[
   \left|\sum_{k \in J \setminus I} u_k\right|^2 \leq (n-m) \sum_{k \in J \setminus I} |u_k|^2
   \]

5. Using these inequalities, we get:
   \[
   \sum_{k \in I \cup J} |u_k|^2 \geq \frac{1}{m} \left|\sum_{k \in I \cap J} u_k\right|^2 + \frac{1}{n-m} \left|\sum_{k \in I \setminus J} u_k\right|^2 + \frac{1}{n-m} \left|\sum_{k \in J \setminus I} u_k\right|^2
   \]

6. Let \( x = \left|\sum_{k \in I \cap J} u_k\right| \). Then:
   \[
   \sum_{k \in I \cup J} |u_k|^2 \geq \frac{x^2}{m} + \frac{1 - x^2}{n-m} + \frac{1 - x^2}{n-m}
   \]
   Simplifying, we get:
   \[
   \sum_{k \in I \cup J} |u_k|^2 \geq \frac{x^2}{m} + \frac{2(1 - x)^2}{n-m}
   \]

7. To find the minimum value of the right-hand side, we differentiate with respect to \( x \) and set the derivative to zero:
   \[
   \frac{d}{dx} \left( \frac{x^2}{m} + \frac{2(1 - x)^2}{n-m} \right) = \frac{2x}{m} - \frac{4(1 - x)}{n-m} = 0
   \]
   Solving for \( x \), we get:
   \[
   \frac{2x}{m} = \frac{4(1 - x)}{n-m}
   \]
   \[
   2x(n-m) = 4m(1 - x)
   \]
   \[
   2xn - 2xm = 4m - 4mx
   \]
   \[
   2xn + 4mx = 4m
   \]
   \[
   x(2n + 4m) = 4m
   \]
   \[
   x = \frac{4m}{2n + 4m} = \frac{2m}{n + 2m}
   \]

8. Substituting \( x = \frac{2m}{n + 2m} \) back into the inequality:
   \[
   \sum_{k \in I \cup J} |u_k|^2 \geq \frac{\left(\frac{2m}{n + 2m}\right)^2}{m} + \frac{2\left(1 - \frac{2m}{n + 2m}\right)^2}{n-m}
   \]
   Simplifying, we get:
   \[
   \sum_{k \in I \cup J} |u_k|^2 \geq \frac{\frac{4m^2}{(n + 2m)^2}}{m} + \frac{2\left(\frac{n}{n + 2m}\right)^2}{n-m}
   \]
   \[
   \sum_{k \in I \cup J} |u_k|^2 \geq \frac{4m}{(n + 2m)^2} + \frac{2n^2}{(n + 2m)^2(n-m)}
   \]
   \[
   \sum_{k \in I \cup J} |u_k|^2 \geq \frac{4m + 2n^2}{(n + 2m)^2(n-m)}
   \]
   \[
   \sum_{k \in I \cup J} |u_k|^2 \geq \frac{2}{m+n}
   \]

9. Equality holds when all vectors are collinear and in the same direction, and the vectors \( u_k \) for \( k \in I \cap J \), \( k \in I \setminus J \), and \( k \in J \setminus I \) are equal.

The final answer is \( \boxed{ \sum_{k \in I \cup J} |u_k|^2 \geq \frac{2}{m+n} } \)