14. (15 points) For 155 boxes containing red, yellow, and blue balls, there are three classification methods: for each color, boxes with the same number of balls of that color are grouped into one class. If all natural numbers from 1 to 30 are the number of boxes in some class of a classification, then,
(1) What is the sum of the number of classes in the three classifications?
(2) Explain that it is possible to find three boxes where the numbers of at least two colors of balls are the same.
【Analysis】According to the problem, we can get $a_{1}, a_{2}, a_{3}, \cdots, a_{i} ; b_{1}, b_{2}, b_{3}, \cdots, b_{j} ; c_{1}, c_{2}, c_{3}, \cdots, c_{k}$, which includes all integers from 1 to 30, so $n \geqslant 30$. On the other hand, 
$$
3 \times 155=a_{1}+a_{2}+a_{3}+\cdots+a_{i}+b_{1}+b_{2}+b_{3}+\cdots+b_{j}+c_{1}+c_{2}+c_{3}+\cdots+c_{k} \geqslant 1+2+3+\cdots+30=\frac{30 \times 31}{2}=465=3 \times 155.
$$
Therefore, the sum of the number of classes in the three categories is 30. This leads to the answer.

【Solution】(1) Since $a_{1}, a_{2}, a_{3}, \cdots, a_{i} ; b_{1}, b_{2}, b_{3}, \cdots, b_{j} ; c_{1}, c_{2}, c_{3}, \cdots, c_{k}$ includes all integers from 1 to 30, so $n \geqslant 30$. On the other hand,
$$
\begin{array}{l}
3 \times 155=a_{1}+a_{2}+a_{3}+\cdots+a_{i}+b_{1}+b_{2}+b_{3}+\cdots+b_{j}+c_{1}+c_{2}+c_{3}+\cdots+c_{k} \geqslant 1+2+3+\cdots+30=\frac{30 \times 31}{2}= \\
465=3 \times 155.
\end{array}
$$

Therefore, $n=i+j+k=30$, the sum of the number of classes in the three categories is 30.
(2) Suppose $a_{1}=30$, and let the class of these 30 boxes be class $A$. Since $i+j+k=30$, there must be $j \leqslant 14$ or $k \leqslant 14$. Without loss of generality, assume $j \leqslant 14$. The 30 boxes of class $A$ are distributed among these no more than 14 classes, so there must be at least one class with at least three boxes, and the number of red balls and yellow balls in these three boxes is the same.