Autumn tournament

(Vlad Spataru) Index the rows and columns from $0$ to $100$. We shall only work modulo $101$ in what follows. Observe that we may let $(0,0)=0$ by adding some constant to all the terms of the table. Next, because of the condition in the statement, $$(u,v)+(i,j)=(u,j)+(i,v),$$ for any $u,v,i,j$. Thus, there exist $a_0=0,a_1,\dots ,a_{100}$ and $b_0=0,b_1,\dots ,b_{100}$ so that $(i,j)=a_i+b_j$. Now, letting ${\Sigma }_a$ and ${\Sigma }_b$ be the sum of the $a_i$ and the $b_j$ respectively, the condition in the statement then yields ${\Sigma }_a+{\Sigma }_b=0$. Note that $$\sum _{i=0}^{100}{x}^{101b_i}={\left(\sum _{i=1}^{100}{x}^{b_i}\right)}^{101}:=\sum _{i=0}^{\infty }{x}^i\alpha (i).$$ Observe that the number of ways of adequately choosing one cell from each row corresponds to the number of $101$-tuples with values from $\{b_0,b_1,\dots ,b_{100}\}$ with sum equal to ${\Sigma }_b$ that is $\alpha ({\Sigma }_b)+\alpha (101+{\Sigma }_b)+\dots$. Using the fact that $101$ is prime and computing the latter expression for $x=\exp (2\pi i/101)$ we get the desired $$0=\sum _{i=0}^{\infty }\alpha (101i+{\Sigma }_b).$$