14. In Greek mythology, the "many-headed serpent" god is composed of some heads and necks, with each neck connecting two heads. With each strike of a sword, one can sever all the necks connected to a certain head $A$, but head $A$ immediately grows new necks connecting to all the heads it was not previously connected to (each head is connected by only one neck). Only by severing the "many-headed serpent" into two disconnected parts can it be defeated. Find the smallest positive integer $N$ such that any "many-headed serpent" god with 100 necks can be defeated with no more than $N$ strikes.
14. Using dots to represent heads and line segments to represent necks, from the given conditions, each line segment connects two points. We call a strike that cuts the necks connected to a head $A$ a "reversal" of vertex $A$. It is easy to see that if a vertex $X$ has a degree no greater than 10, then by performing a "reversal" on all vertices adjacent to $X$, vertex $X$ can be isolated (i.e., $X$ will no longer be connected to any other vertices). If a vertex $X$ is not adjacent to at most $n (n \leqslant 9)$ vertices, then by first performing a "reversal" on $X$ and then performing a "reversal" on each of these $n$ vertices, vertex $X$ can be isolated.

If each vertex has at least 11 adjacent vertices and at least 10 non-adjacent vertices, then there must be at least 22 vertices in total. Thus, the number of line segments (necks) in the graph is no less than $\frac{1}{2} \times 22 \times 11 > 100$, which does not meet the problem's conditions. Therefore, it is sufficient to cut 10 times to meet the requirements. Furthermore, we provide an example to show that cutting 9 times is insufficient. Suppose there are two groups of 10 heads each, and each head in one group is connected to every head in the other group, resulting in exactly 100 necks. If only 9 cuts are made, then at least one head in each group will not be struck, let these be $A$ and $B$. For the remaining 18 heads, by the problem's conditions, any head $C$ will always be connected to exactly one of $A$ or $B$ and not the other, and since $A$ and $B$ are connected, the "many-headed snake" remains connected.