(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
$$0 = \sum_{i=0}^{\infty} \alpha (101i + \Sigma_b).$$