6. Assign 18 table tennis players numbered $1, 2, \cdots, 18$ to 9 tables for singles matches, with the rule that the sum of the numbers of the two players on each table must be a perfect square greater than 4. Let $\{7$ and 18 play against each other $\}$ be event $p$. Then $p$ is $(\cdots)$.
(A) impossible event
(B) random event with probability $\frac{1}{17}$
(C) random event with probability $\frac{1}{3}$
(D) certain event
6.D.

Since the sum of the two largest numbers is $18+17=35$ $<36$, the sum of the numbers of two players on the same table can only be 3 square numbers: $25, 16, 9$. Let the number of pairs with sums of $25, 16, 9$ be $x, y, z$ respectively $(x, y, z$ are non-negative integers $)$. According to the problem, we have
$$
\left\{\begin{array}{l}
25 x+16 y+9 z=1+2+\cdots+18, \\
x+y+z=9, \\
x \geqslant 0, y \geqslant 0, z \geqslant 0,
\end{array}\right.
$$

which simplifies to $\left\{\begin{array}{l}16 x+7 y+9(x+y+z)=171, \\ x+y+z=9, \\ x \geqslant 0, y \geqslant 0, z \geqslant 0 .\end{array}\right.$
resulting in $\left\{\begin{array}{l}16 x+7 y=90, \\ x \geqslant 0, y \geqslant 0, z \geqslant 0 .\end{array}\right.$
Given $0 \leqslant x \leqslant \frac{90}{16}<6$, $x$ can only take non-negative integers $0,1,2,3,4,5$. Substituting each value, we find the unique non-negative integer solution $x=3, y=6, z=0$.
Next, we discuss the match-ups on the 9 tables.
(1) Since $x=3$, the square number 25 can only appear 3 times. The 3 players with numbers not less than 16, $18,17,16$, can only correspond to the square number 25. Therefore, the pairs with a sum of 25 are: 18 vs 7, 17 vs 8, 16 vs 9.
(2) Since $y=6$, after removing $18,17,16,9,8,7$, the remaining 12 players can only correspond to the square number 16, with the pairs: 1 vs 15, 2 vs 14, 3 vs 13, 4 vs 12, 5 vs 11, 6 vs 10.

Thus, the arrangement is possible and the solution is unique. The match-ups on the 9 tables are:
$(18,7),(17,8),(16,9),(15,1),(14,2)$, $(13,3),(12,4),(11,5),(10,6)$.
Event $p$ is a certain event.