14. For 155 boxes containing balls of three colors: red, yellow, and blue, 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 balls of at least two colors are the same.
Let the first, second, and third types of classification be divided into $i, j, k$ categories, respectively, with the number of boxes in each category being
$$
a_{1}, a_{2}, a_{3}, \cdots, a_{i} ; b_{1}, b_{2}, b_{3}, \cdots, b_{j} ; c_{1}, c_{2}, c_{3}, \cdots, c_{k}
$$

Let $n=i+j+k$
1) Since $a_{1}, a_{2}, a_{3}, \cdots, a_{i} ; b_{1}, b_{2}, b_{3}, \cdots, b_{j} ; c_{1}, c_{2}, c_{3}, \cdots, c_{k}$ include all integers from 1 to 30, we have $\mathrm{n} \geqslant 30$. On the other hand,
$$
\begin{array}{l}
3 \times 155=\mathrm{a}_{1}+\mathrm{a}_{2}+\mathrm{a}_{3}+\cdots+\mathrm{a}_{\mathrm{i}}+\mathrm{b}_{1}+\mathrm{b}_{2}+\mathrm{b}_{3}+\cdots+\mathrm{b}_{\mathrm{j}}+\mathrm{c}_{1}+\mathrm{c}_{2}+\mathrm{c}_{3}+\cdots+\mathrm{c}_{\mathrm{k}} \\
\geqslant 1+2+3+\cdots+30=\frac{30 \times 31}{2}=465=3 \times 155
\end{array}
$$

Therefore, $n=i+j+k=30$, the sum of the number of categories in the three types of classification is 30.
2) Without loss of generality, assume $a_{1}=30$, and let the 30 boxes in this category be of type $A$. 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 type $A$ are distributed among these no more than 14 categories, so there must be at least one category with at least three boxes, and the number of red balls and yellow balls in these three boxes is the same.