Note that half the numbers plus one are selected.
Let $j$ be such a positive integer. We can now divide the set into $j$ sets $J_i$ of equal size (because $j$ divides $4n$), determined by their remainder after division by $j$. I.e.
$$J_i = \{ n | n \cong i \mod j \}$$
with $i$ ranging from $0$ to $j - 1$.
Each of them contains $2\cdot \frac{2n}{j}$ numbers, which is an even number.
As we have picked $2n + 1$ numbers, through the pigeonhole principle, over half of the numbers in (at least) one of the $J_i$ have been selected. Two of them must have difference $j$. Why?
Rearrange the integers from $1$ to $4n$ into
$$\begin{equation}\begin{aligned}
\{& (1,j+1), (2, j + 2), (3, j + 3), \, \ldots \, , (j, 2j), \\
& (2j + 1, 3j + 1), (2j + 2, 3j + 2), \, \ldots \, , (3j, 4j), \\
& \vdots \\
& (4n - 2j + 1, 4n - j + 1), (4n - 2j + 2, 4n - j + 2), \, \ldots \, , (4n - j, 4n) \}
\end{aligned}\end{equation}\tag{1}\label{eq1A}$$
Note each member of the set above is a pair of integers with a difference of $j$. Also, since $j \mid 2n \implies 2j \mid 4n$, all of the integers from $1$ to $4n$ are in this set exactly once each. Finally, this gives that the number of elements of the set is the $4n$ integers divided by the $2$ integers in each pair, i.e., $\frac{4n}{2} = 2n$. Alternatively, you also can get the number of elements by multiplying the number of columns times the number of rows, i.e.,
$$j \times \left(\frac{4n-2j}{2j} + 1 \right) = j \times \left(\frac{2n}j - 1 + 1\right) = 2n \tag{2}\label{eq2A}$$
Thus, by the Pigeonhole principle, choosing $2n + 1$ integers between $1$ and $4n$ means that both of the integers from at least one of the element pairs in \eqref{eq1A} must be chosen, with these $2$ integers therefore having a difference of $j$.
Best Answer
The set of all integers between $1$ and $100$ can be written as $$\lbrace 1,..., 24 \rbrace \cup \lbrace 25, ..., 48 \rbrace \cup \lbrace 49,..., 72\rbrace \cup \lbrace 73, ..., 96 \rbrace \cup \lbrace 97,..., 100 \rbrace $$
If you choose $55$ numbers, then at least $51$ of them are taken in the first four sets. So at least $13$ of them are taken in one of the four sets. Now you can conclude.