$4 \cdot 180$ Try to determine all the quadruples $\left(p_{1}, p_{2}, p_{3}, p_{4}\right)$, where $p_{1}, p_{2}$, $p_{3}, p_{4}$ are all prime numbers, and satisfy:
(1) $p_{1}<p_{2}<p_{3}<p_{4}$;
(2) $p_{1} p_{2}+p_{2} p_{3}+p_{3} p_{4}+p_{4} p_{1}=882$.
[Solution] Rewrite condition (2) as
$$
\left(p_{1}+p_{3}\right)\left(p_{2}+p_{4}\right)=2 \times 3^{2} \times 7^{2},
$$

The right-hand side of this equation cannot be divisible by 4, so one of the factors on the left must be odd, thus
$$
p_{1}=2, p_{1}+p_{3} \text { is an odd factor of } 882.
$$

Since $p_{1}+p_{3}<p_{2}+p_{4}$, we have
$$
p_{1}+p_{3}<\sqrt{882}<30,
$$

Therefore,
$$
\begin{aligned}
2+p_{3} & \in\{3,7,9,21\}, \\
p_{3} & \in\{5,7,19\}.
\end{aligned}
$$

If $p_{3}=5$, then $p_{2}=3, p_{2}+p_{4}=2 \times 3^{2} \times 7$, $p_{4}=2 \times 3^{2} \times 7-3=3 \times(2 \times 3 \times 7-1)=3 \times 41$ is not a prime, contradiction.

If $p_{3}=7$, then $p_{2}+p_{4}=2 \times 7^{2}=98, p_{2}=3$ or $5, p_{4}=95$ or 93 are not primes, contradiction.
If $p_{3}=19$, then $p_{2}+p_{4}=2 \times 3 \times 7=42,3 \leqslant p_{2} \leqslant 17$.
When $p_{2}=3$, $p_{4}=39$ is not a prime, contradiction.
When $p_{2}=5$, $p_{4}=37$.
When $p_{2}=7$, $p_{4}=35$ is not a prime, contradiction.
When $p_{2}=11$, $p_{4}=31$.
When $p_{2}=13$, $p_{4}=29$.
When $p_{2}=17$, $p_{4}=25$ is not a prime, contradiction.
Therefore, the quadruples that satisfy the conditions are
$$
(2,5,19,37),(2,11,19,31),(2,13,19,29).
$$