Let $a$, $b$, $c$, $a+b-c$, $a+c-b$, $b+c-a$, and $a+b+c$ be 7 distinct prime numbers, and suppose that the sum of two of $a$, $b$, and $c$ is 800. Let $d$ be the difference between the largest and smallest of these 7 prime numbers. Find the maximum possible value of $d$.
(Liang Darong, problem contributor)
$$
\begin{array}{l}
\text { Let's assume } a<b<c<d \text {, and } a, b, c, d \text { are all prime numbers. } \\
\therefore c<a+b<a+c<b+c .
\end{array}
$$

Also, since one of $a+b$, $a+c$, and $b+c$ is 800,
$$
\therefore c<800 \text {. }
$$

Since $799=17 \times 47$ and 798 are not prime numbers, but 797 is a prime number, we have
$$
\begin{array}{l}
c \leqslant 797, d \geqslant 1594 . \\
\text { In another case, when } a+b=1300 \text {, note that } \\
a=5, b=795, \\
a=7, b=793=13 \times 61, \\
a=11, b=789=3 \times 263
\end{array}
$$

are not all prime numbers, thus cannot meet the requirements of the problem.
However, when $a=13, b=787$, both are prime numbers, and at this time, $a+b-c=3, a+c-b=23$ are also prime numbers. It is easy to verify that $b+c-a=1571$ and $a+b+c=1597$ are also prime numbers.
In summary, the maximum possible value of $d$ is 1594.