Let $n$ be a positive integer. For each $4n$-tuple of nonnegative real numbers $a_1,\ldots,a_{2n}$, $b_1,\ldots,b_{2n}$ that satisfy $\sum_{i=1}^{2n}a_i=\sum_{j=1}^{2n}b_j=n$, define the sets
\[A:=\left\{\sum_{j=1}^{2n}\frac{a_ib_j}{a_ib_j+1}:i\in\{1,\ldots,2n\} \textup{ s.t. }\sum_{j=1}^{2n}\frac{a_ib_j}{a_ib_j+1}\neq 0\right\},\]
\[B:=\left\{\sum_{i=1}^{2n}\frac{a_ib_j}{a_ib_j+1}:j\in\{1,\ldots,2n\} \textup{ s.t. }\sum_{i=1}^{2n}\frac{a_ib_j}{a_ib_j+1}\neq 0\right\}.\]
Let $m$ be the minimum element of $A\cup B$. Determine the maximum value of $m$ among those derived from all such $4n$-tuples $a_1,\ldots,a_{2n},b_1,\ldots,b_{2n}$.

[I]Proposed by usjl.[/i]
1. **Define the problem and constraints:**
   We are given two sets of nonnegative real numbers \(a_1, a_2, \ldots, a_{2n}\) and \(b_1, b_2, \ldots, b_{2n}\) such that:
   \[
   \sum_{i=1}^{2n} a_i = \sum_{j=1}^{2n} b_j = n
   \]
   We need to define the sets \(A\) and \(B\) as follows:
   \[
   A := \left\{\sum_{j=1}^{2n} \frac{a_i b_j}{a_i b_j + 1} : i \in \{1, \ldots, 2n\} \text{ s.t. } \sum_{j=1}^{2n} \frac{a_i b_j}{a_i b_j + 1} \neq 0 \right\}
   \]
   \[
   B := \left\{\sum_{i=1}^{2n} \frac{a_i b_j}{a_i b_j + 1} : j \in \{1, \ldots, 2n\} \text{ s.t. } \sum_{i=1}^{2n} \frac{a_i b_j}{a_i b_j + 1} \neq 0 \right\}
   \]
   We need to find the maximum value of the minimum element \(m\) of \(A \cup B\).

2. **Simplify the problem using permutations:**
   Without loss of generality, we can permute \(a_i\) and \(b_j\) such that \(a_i = 0\) for \(i > x\) and \(b_j = 0\) for \(j > y\), where \(x\) and \(y\) are the number of nonzero terms in \(a_i\) and \(b_j\) respectively.

3. **Apply Cauchy-Schwarz inequality:**
   Consider the sum:
   \[
   \sum_{j=1}^y \frac{a_i b_j}{a_i b_j + 1} = y - \sum_{j=1}^y \frac{1}{a_i b_j + 1}
   \]
   Using the Cauchy-Schwarz inequality, we get:
   \[
   \sum_{j=1}^y \frac{1}{a_i b_j + 1} \geq \frac{(\sum_{j=1}^y 1)^2}{\sum_{j=1}^y (a_i b_j + 1)} = \frac{y^2}{y + n a_i}
   \]
   Therefore:
   \[
   \sum_{j=1}^y \frac{a_i b_j}{a_i b_j + 1} \leq y - \frac{y^2}{y + n a_i}
   \]

4. **Define \(A_i\) and bound the sum:**
   Let:
   \[
   A_i = \sum_{j=1}^y \frac{a_i b_j}{a_i b_j + 1}
   \]
   Then:
   \[
   \sum_{i=1}^x A_i \leq xy - y^2 \sum_{i=1}^x \frac{1}{y + n a_i}
   \]
   Using Cauchy-Schwarz again:
   \[
   \sum_{i=1}^x \frac{1}{y + n a_i} \geq \frac{(\sum_{i=1}^x 1)^2}{\sum_{i=1}^x (y + n a_i)} = \frac{x^2}{xy + n^2}
   \]
   Therefore:
   \[
   \sum_{i=1}^x A_i \leq xy - y^2 \frac{x^2}{xy + n^2} = \frac{x y n^2}{xy + n^2}
   \]

5. **Apply the Pigeonhole Principle:**
   By the Pigeonhole Principle, at least one \(A_i\) satisfies:
   \[
   0 < A_i \leq \frac{y n^2}{xy + n^2}
   \]

6. **Maximize the minimum of \(A \cup B\):**
   The minimum of \(A \cup B\) is:
   \[
   \frac{x y n^2}{(xy + n^2) \times \max \{x, y\}}
   \]
   Without loss of generality, assume \(x \leq y\). Then:
   \[
   \frac{x y n^2}{(xy + n^2) \times y} = \frac{x n^2}{xy + n^2} \leq \frac{x n^2}{x^2 + n^2} \leq \frac{n}{2}
   \]
   Equality holds when \(x = y = n\).

Conclusion:
The maximum value of \(m\) is \(\frac{n}{2}\).

The final answer is \(\boxed{\frac{n}{2}}\)