4. We refer to both grandfather and maternal grandfather as grandfather, and both grandson and maternal grandson as grandson. 20 children attend a party, and it is known that any two of these children have a common grandfather. How many grandchildren does the grandfather with the most grandchildren have at least?
4. Let $b_{1}, b_{2}, \cdots, b_{20}$ represent the children, and $G_{1}, G_{2}, \cdots, G_{m}$ represent the grandfathers. Construct a $20 \times m$ table, where the value in the $i$-th row and $j$-th column, $l_{j}=\sum_{i=1}^{20} x_{i j}$, represents the number of grandsons of $G_{j}$. Since each child has exactly 2 grandfathers, $r_{i}=2(1 \leqslant i \leqslant 20)$. Therefore, $\sum_{j=1}^{m} l_{j}=\sum_{j=1}^{m} \sum_{i=1}^{20} x_{i j}=\sum_{i=1}^{20} \sum_{j=1}^{m} x_{i j}=\sum_{i=1}^{20} r_{i}=40$ (1). Assume $l_{1}=\max \left\{l_{1}, l_{2}, \cdots, l_{m}\right\}$. If all the numbers in the first column are 1, then $l_{1}=20$. Now, assume not all numbers in the first column are 1, and let $x_{21}=0$. Since $b_{2}$ and $b_{1}$ have a common grandfather, but $G_{1}$ is not the grandfather of $b_{2}$, assume the common grandfather of $b_{2}$ and $b_{1}$ is $G_{2}$, and the other grandfather of $b_{2}$ is $G_{3}$. If every grandson of $G_{1}$ is also a grandson of $G_{2}$, then $G_{2}$ would have at least one more grandson, $b_{2}$, than $G_{1}$, so $l_{2} \geqslant l_{1}+1$, which contradicts the assumption that $l_{1}$ is the maximum. Therefore, $G_{1}$ must have a grandson (let's say $b_{3}$) who is not a grandson of $G_{2}$, i.e., $x_{31}=1, x_{32}=0$. However, $b_{3}$ and $b_{2}$ have a common grandfather, which cannot be $G_{2}$ (since $x_{32}=0$), so it must be $G_{3}$, i.e., $x_{33}=1$.

Next, we prove that for any $b_{i}(4 \leqslant i \leqslant 20)$, his grandfathers can only be two of $G_{1}, G_{2}, G_{3}$, thus $m=3$. In fact, for $b_{i}(i \geqslant 4)$, the common grandfather with $b_{1}$ can only be $G_{1}$ or $G_{2}$, the common grandfather with $b_{2}$ can only be $G_{2}$ or $G_{3}$, and the common grandfather with $b_{3}$ can only be $G_{1}$ or $G_{3}$. Therefore, the grandfathers of $b_{i}(i \geqslant 4)$ can only be two of $G_{1}, G_{2}, G_{3}$. Thus, $m=3$. From (1), we get $l_{1} \geqslant \frac{1}{3}\left(l_{1}+l_{2}+l_{3}\right)=\frac{40}{3}$, so $l_{1} \geqslant 14$.

On the other hand, suppose the grandsons of $G_{1}$ are $b_{1}, b_{2}, \cdots, b_{14}$, the grandsons of $G_{2}$ are $b_{1}, \cdots, b_{7}$ and $b_{15}, \cdots, b_{20}$, and the grandsons of $G_{3}$ are $b_{8}, b_{9}, \cdots, b_{20}$. Then any two grandsons have a common grandfather. Among them, $G_{1}$ has the most grandsons, with 14. In conclusion, the grandfather with the most grandsons has at least 14 grandsons.