For any set of $L+2 (L<78)$ integers there exists two integers $x$ and $y$ such that $x + y$ or $x - y$ is divisible by $2L$. To show that for any set of $L+2 (L<78)$ integers there exists two integers $x$ and $y$ such that $x + y$ or $x - y$ is divisible by $2L$.

For any integer $a$ we have $$a \equiv 0, \pm 1, \pm2, \cdots,\pm (L-1),L \mod 2L$$ thus
$$a^2 \equiv 0,1,4,\cdots,L^2 \mod 2L.$$ Hence there are a set of $L+1$ in-congruent squares $\mod 2L$.
Thus from any set of $L+2$ integers there exists at least $2$ integers such that
$$x^2 \equiv y^2 \mod 2L$$ thus $2L$ divides $x^2 - y^2 =(x-y)(x+y)$.
Note that $x+y + x-y= 2x$ hence both of them have the same parity and since $2$ divides them both are even.

But from here how to conclude that $2L$ divides $x+y$ or $x-y$?
 A: Assume that you have $(L+2)$ integers, and that it is not the case that there are any $2$ integers $x,y$ from this group such that $(2L)$ divides $(x - y)$.
For each one of the $(L+2)$ integers, focus on its congruence class, $\pmod{2L}$.  Since it is assumed that there is no occurrence of a pair $x,y$ such that $(2L)$ divides $(x - y)$ it must be the case that the $(L+2)$ integers represent $(L+2)$ distinct congruence classes, from the set of all possible congruence classes, $\pmod{2L}$.
Clearly, the set of these congruence classes may be represented by the elements in the set $S = \{0,1,2,\cdots, (2L-1)\}.$
The sole issue, is whether it is possible to take any subset $T$ from set $S$, such that $T$ has exactly $(L+2)$ distinct elements and have it be the case that there is no pair $x,y$ from this set $T$ such that $(2L)$ divides $(x + y)$.
Answer: No because:
Each time that you take an element from the set $S$ and add it to the subset $T$, you are eliminating a $2$nd element from the set $S$.  Otherwise, you would have to have the satisfying pair of elements $x,y$ in $T$.
The exception are the elements $0$ and $L$.  Setting $L = 5$ will provide illustration.
From each of the following ordered pairs, you can only take one element, to add to the subset $T$:
$(1,9), (2,8), (3,7), (4,6)$.
This makes $(5 - 1)$ elements.
Then, you can also have the elements $0$ and $(5)$.
This makes $(5 + 1)$ elements.
Then, when you add the $7$th element, you have to end up with some pair of elements, from 
$(1,9), (2,8), (3,7), (4,6)$
such that both of the elements in this pair are in the set $T$.
For the general value of $L$, the analysis is identical.
That is, you have the $(L-1)$ pairs
$(1,[2L-1]), (2,[2L-2]), \cdots, (L-1,L+1)$.
Then, you have the elements $0$ and $L$.
This makes $L+1$ elements.  When you go to add one extra element, the result has to be that one of the pairs:
$(1,[2L-1]), (2,[2L-2]), \cdots, (L-1,L+1)$
has both of its elements in the subset $T$.
