Three. (25 points) 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. Can this rule be implemented? If the rule cannot be implemented, provide a proof; if the rule can be implemented, determine whether the implementation is unique.
Three, Solution 1: 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 one of the three square numbers: $25, 16, 9$. Let the number of pairs with sums of $25, 16, 9$ be $x$, $y$, and $z$ 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 is equivalent to $\left\{\begin{array}{l}16 x+7 y+9(x+y+z)=171, \\ x+y+z=9, \\ x \geqslant 0, y \geqslant 0, z \geqslant 0,\end{array}\right.$
leading to $\left\{\begin{array}{l}16 x+7 y=90, \\ x \geqslant 0, y \geqslant 0, z \geqslant 0 .\end{array}\right.$
Since $0 \leqslant x \leqslant \frac{90}{16}<6$, $x$ can only take the non-negative integers $0,1,2,3,4,5$. Substituting each value for inspection, we get the unique non-negative integer solution
$$
x=3, y=6, z=0 .
$$

Now let's 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 must be: 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:
$$
(18,7),(17,8),(16,9),(15,1),(14,2) \text {, }
$$
$(13,3),(12,4),(11,5),(10,6)$.
Solution 2: 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 one of the three square numbers: $25,16,9$.
(1) The 3 players with numbers not less than 16 can only correspond to the square number 25, with the pairs: 18 vs 7, 17 vs 8, 16 vs 9.
(2) After removing $18,17,16,9,8,7$, among the remaining 12 players: Since 1 cannot form the square number 9 with 8, 1 can only be paired with 15; Since 2 cannot form the square number 9 with 7, 2 can only be paired with 14.
(3) After removing $15,14,2,1$, among the remaining 8 players: Since 10 cannot form the square number 25 with 15, 10 can only be paired with 6; Since 11 cannot form the square number 25 with 14, 11 can only be paired with 5.
(4) Among the last 4 numbers $3,4,12,13$, the pairs that can form a square number are: 13 vs 3, 12 vs 4.

Thus, the arrangement is possible, and the solution is unique. The match-ups on the 9 tables are:
$$
\begin{array}{l}
(18,7),(17,8),(16,9),(15,1),(14,2) \\
(13,3),(12,4),(11,5),(10,6)
\end{array}
$$