The seventy-third question: Let $M$ be the set of permutations $\left(a_{1}, a_{2}, \ldots, a_{10}\right)$ of the set $\{1,2, \ldots, 10\}$ that satisfy the following conditions: $3 k \mid\left(a_{1}+a_{2}+\ldots+a_{3 k}\right)(k=1, 2, 3)$. Find the number of elements in the set $M$.

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The 73rd question:
Solution: Let $S_{n}=\sum_{i=1}^{n} a_{i}$, then $S_{10}=55$, so $a_{10} \equiv 55-S_{9} \equiv 1 \pmod{9}$, hence $a_{10}=10$ or 1.
(1) If $a_{10}=10$, then it is only necessary to ensure that $3 \mid S_{3}$ and $6 \mid S_{6}$.
First, consider $6 \mid S_{6}$. Classify the numbers in the set $M=\{1,2, \ldots, 9\}$ by their remainders modulo 6: $A_{1}=\{1,7\}$, $A_{2}=\{2,8\}$, $A_{3}=\{3,9\}$, $A_{4}=\{4\}$, $A_{5}=\{5\}$, $A_{6}=\{6\}$. Next, consider the permutations $\left(a_{1}, a_{2}, \ldots, a_{6}\right)$ that satisfy $6 \mid S_{6}$.
(1) The six numbers in the set $A_{1} \cup A_{4} \cup A_{3} \cup A_{6}$ clearly satisfy the requirement. Denote the number with a remainder of $r$ modulo 6 as $r$, then the six numbers are $1, 1, 3, 3, 4, 0$.
(2) The six numbers in the set $A_{1} \cup A_{2} \cup A_{3}=\{1,7,2,8,3,9\}$ are denoted as $1, 1, 2, 2, 3, 3$.
(3) Four numbers from the set $A_{1} \cup A_{5} \cup A_{6}$ and one number each from $A_{2}$ and $A_{3}$, denoted as $1, 1, 2, 3, 5, 0$.
(4) Four numbers from the set $A_{2} \cup A_{4} \cup A_{6}$ and one number each from $A_{1}$ and $A_{3}$, denoted as $1, 2, 2, 3, 4, 0$.
(5) Four numbers from the set $A_{3} \cup A_{4} \cup A_{5}$ and one number each from $A_{1}$ and $A_{2}$, denoted as $1, 2, 3, 3, 4, 5$.
Then, classify the above five groups of numbers by their remainders modulo 3. Similarly, the first group of numbers is denoted as $1, 1, 1, 0, 0, 0$, and the other four groups of numbers can all be denoted as $1, 1, 2, 2, 0, 0$.

Therefore, the number of permutations that satisfy the problem's conditions formed by the first group of numbers is $\mathrm{C}_{2}^{1} \mathrm{~A}_{3}^{3} \mathrm{~A}_{3}^{3}$; the number of permutations that satisfy the problem's conditions formed by the second group of numbers is $\mathrm{C}_{2}^{1} \mathrm{C}_{2}^{1} \mathrm{C}_{2}^{1} \mathrm{~A}_{3}^{3} \mathrm{~A}_{3}^{3}$; the number of permutations that satisfy the problem's conditions formed by the third, fourth, and fifth groups of numbers is $\mathrm{C}_{2}^{1} \mathrm{C}_{2}^{1} \mathrm{C}_{2}^{1} \mathrm{C}_{2}^{1} \mathrm{C}_{2}^{1} \mathrm{~A}_{3}^{3} \mathrm{~A}_{3}^{3}$.
In summary, when $a_{10}=10$, the number of permutations that satisfy the problem's conditions is
$$
\mathrm{A}_{3}^{3}\left(\mathrm{C}_{2}^{1} \mathrm{~A}_{3}^{3} \mathrm{~A}_{3}^{3}+\mathrm{C}_{2}^{1} \mathrm{C}_{2}^{1} \mathrm{C}_{2}^{1} \mathrm{~A}_{3}^{3} \mathrm{~A}_{3}^{3}+3 \times \mathrm{C}_{2}^{1} \mathrm{C}_{2}^{1} \mathrm{C}_{2}^{1} \mathrm{C}_{2}^{1} \mathrm{C}_{2}^{1} \mathrm{~A}_{3}^{3} \mathrm{~A}_{3}^{3}\right)=22896
$$
(2) When $a_{10}=1$, let $a_{i}=a_{i}^{\prime}+1$, then $3 k\left|\left(\sum_{i=1}^{3 k} a_{i}\right) \Leftrightarrow 3 k\right|\left(\sum_{i=1}^{3 k} a_{i}^{\prime}\right)(k=1, 2, 3)$. Therefore, when $a_{10}=1$, the number of permutations that satisfy the problem's conditions is also 22896.
Combining (1) and (2), the number of elements in the set $M$ is $22896 \times 2=45792$.