Let the given row be $a_1, \dots, a_{51}$. Take a circle $\gamma$ of length 100. Mark 51 blue points on it so that they determine 51 consecutive arcs of lengths $a_1, \dots, a_{51}$. One may imagine each number $a_i$ written next to an arc of $\gamma$ with length $a_i$. So, starting from a certain blue point $B$, the numbers are arranged around $\gamma$ as in the given row.

The claim that $k$ or $100-k$ is representable is equivalent to saying that there is an arc $\alpha$ of length $k$ with blue endpoints. Indeed consider $\alpha$ and its complementary arc $\alpha'$, of length $100-k$. Point $B$ cannot be interior to both $\alpha$ and $\alpha'$, hence $k$ or $100-k$ equals the sum of several consecutive numbers in the initial row.

We show that for each $k \le 50$ there is an arc of length $k$ with blue endpoints (once this is proven, the case $k > 50$ follows trivially). Mark 49 more red points on $\gamma$ so that the total of marked 100 points, blue and red, yields a division into arcs of length 1. Now $k=50$ is almost immediate. To each blue point assign its diametrically opposite point (the number 100 of division points is even). The 51 assigned points are distinct. Since there are $49 < 51$ red points, some assigned point is blue, which is enough.

The essential case $k < 50$ needs an appropriate modification. For each blue point $X$ write down the endpoints of the arc with length $2k$ and midpoint $X$ (this arc is "proper", it does not overlap itself). One obtains a list of $2 \cdot 51 = 102$ points. It is clear that no point occurs in it more than twice.

Therefore the list contains at least 51 distinct points. There are only $49 < 51$ red points. Hence some blue point is listed, completing the argument.