Suppose these 65 couples have already chosen an arrangement of carrying satisfying the conditions of the problem. We prove that there must be a couple on the same motorcycle (in fact there must be exactly one such couple), so that after removing them, the remaining 64 couples can simply keep using the original arrangement, which clearly satisfies the condition of the problem.

Number all the couples from 1 to 65, and define the function $f(a) = b$ to mean that girl number $a$ is carried by boy number $b$. Then $f(a)$ is a one-to-one and onto function, so if for all $1 \le a \le 65$ we keep substituting $a$ into $f$ until the value returns to $a$, and write the process as a cycle, we obtain $a \rightarrow f(a) \rightarrow f(f(a)) \cdots \rightarrow f^{(k)}(a) = f(f(\cdots f(a)\cdots)) = a$; in this way, 1 through 65 can be split into a number of disjoint cycles, so there must be a cycle with an odd number of elements. Suppose this cycle is
$$a_1 \rightarrow a_2 \rightarrow \cdots \rightarrow a_k \rightarrow a_1$$
where $k$ is odd.

Suppose $k > 1$; we define the symbol $g(a, b)$ to mean that boy number $a$ and boy number $b$ know each other, and $<a, b>$ to mean that boy number $a$ and boy number $b$ do not know each other. Suppose we have $(a_1, a_2)$; then since $f(a_1) = a_2, f(a_2) = a_3$, we know that the boyfriends of the girls carried by boy $a_2$ and boy $a_3$ (namely $a_1$ and $a_2$) know each other, and so we obtain $<a_2, a_3>$; similarly, from $f(a_2) = a_3, f(a_3) = a_4$ we know that the boyfriends of the girls carried by boy $a_3$ and boy $a_4$ (namely $a_2$ and $a_3$) do not know each other, so boy $a_3$ and boy $a_4$ should know each other, giving us $(a_3, a_4)$; continuing this reasoning gives:
$$(a_1, a_2) \rightarrow <a_2, a_3> \rightarrow (a_3, a_4) \cdots \rightarrow (a_k, a_1) \rightarrow <a_1, a_2>$$
a contradiction! If instead the case $<a_1, a_2>$ held, the same reasoning can still be used to derive a contradiction!

Therefore $k=1$, that is $f(a_1) = a_1$, which completes the proof.