Let $a_1, a_2, \dots, a_{51}$ be the 51 numbers with $a_1 + a_2 + \dots + a_{51} = 100$. On a circle of total length 100 we place 100 points such that the length of the arc between any two neighboring points (of these 100) is equal to 1. We fix one of these points and denote it by $A_1$. Then, we mark on the circle points $A_2, A_3, \dots, A_{51}$, in this order, such that the length of each arc $A_iA_{i+1}$ is $a_i$ for all $i = 1, 50$. Clearly, the length of the arc $A_{51}A_1$ will be $a_{51}$.

We color the points $A_1, A_2, \dots, A_{50}$ with blue, and the other 49 points with red. We prove that for all $k \in [1, 99]$ we can find two blue points, $A_i$ and $A_j$, such that the lengths of the two arcs determined by these two points are $k$ and $100-k$. It is sufficient to prove this statement for $k \le 50$.

For $k = 50$ we have 51 blue points, so there must be a pair of blue points that are diametrically opposed. For $k < 50$, consider for each blue point $A_j$ the points $B$ such that the length of the arc $A_jB$ is $k$. There are two such points for each point $A_j$. If any of these points is blue, we have found an arc $A_iA_j$ of length $k$. Assume all these points $B$ are red; each of these points $B$ has been considered at most twice, but each blue point uses two red ones. Thus, 51 blue points require 51 red ones, but only 49 are available. We conclude that our assumption was false, therefore an arc $A_iA_j$ of length $k$ must exist.

Now we cut the circle at $A_1$ and turn it into a line segment. For all $k$, we might have thus cut one of the arcs $A_iA_j$, either the one of length $k$ or the one of length $100-k$, but not both. Thus, for all $k$, either a line segment $A_iA_j$ of length $k$, or one of length $100-k$, must exist.