2. Find all quadruples of positive real numbers $(a, b, c, d)$ such that
$$
\begin{array}{l}
\text { satisfy } a+b+c+d=1, \text { and } \\
\max \left\{\frac{a^{2}}{b}, \frac{b^{2}}{a}\right\} \max \left\{\frac{c^{2}}{d}, \frac{d^{2}}{c}\right\} \\
=(\min \{a+b, c+d\})^{4} .
\end{array}
$$
2. Assume $a+b \leqslant c+d$.

Notice that, swapping $a$ with $b$ and $c$ with $d$ does not change the condition of the problem.
Without loss of generality, assume $a \leqslant b, c \leqslant d$.
From $a^{2} \leqslant b^{2} \Rightarrow \frac{a^{2}}{b} \leqslant \frac{b^{2}}{a}$.
Similarly, $\frac{c^{2}}{d} \leqslant \frac{d^{2}}{c}$.
Thus, $\frac{b^{2}}{a} \cdot \frac{d^{2}}{c}=(a+b)^{4}$.
Since $\frac{b^{2}}{a} \geqslant b, \frac{d^{2}}{c} \geqslant d$, we have $\frac{b^{2}}{a} \cdot \frac{d^{2}}{c} \geqslant b d$.
We now prove: $(a+b)^{4} \leqslant b d$.
From $a \leqslant b \Rightarrow b \geqslant \frac{1}{2}(a+b)$.
Similarly, $d \geqslant \frac{1}{2}(c+d)$.
Since $c+d \geqslant a+b$, we have $d \geqslant \frac{1}{2}(a+b)$.
Given $a+b+c+d=1$ and $a+b \leqslant c+d$
$\Rightarrow a+b \leqslant \frac{1}{2}$.
Then $b d \geqslant \frac{1}{4}(a+b)^{2}$
$\geqslant(a+b)^{2}(a+b)^{2}=(a+b)^{4}$.
Thus, $(a+b)^{4}=b d$.
It is easy to verify that when $a=b=c=d=\frac{1}{4}$, the equality holds.
Therefore, $(a, b, c, d)=\left(\frac{1}{4}, \frac{1}{4}, \frac{1}{4}, \frac{1}{4}\right)$.