India — Team Selection Test, 2010

Consider the first column. Let $M$ be the maximum value of the entries in this column. Then $M \ge 1/10$, by pigeonhole principle. Writing the first column $a_{1,1}, a_{2,1}, \dots, a_{10,1}$, let $M = a_{j,1}$, for some $j$. Consider the sum
$$\sum_{l=2}^{10} (a_{j,1}a_{k,l} + a_{j,l}a_{k,1}),$$
where $k \neq j$. This is equal to
$$a_{j,1} \left( \sum_{l=2}^{10} a_{k,l} \right) + a_{k,1} \left( \sum_{l=2}^{10} a_{j,l} \right) = a_{j,1} (1 - a_{k,1}) + a_{k,1} (1 - a_{j,1}).$$
Varying $k$ from $1$ to $10$, $k \neq j$, we get
$$\begin{align*} \sum_{k \neq j} \sum_{l=2}^{10} (a_{j,1}a_{k,l} + a_{j,l}a_{k,1}) &= \sum_{k \neq j} a_{j,1}(1-a_{k,1}) + a_{k,1}(1-a_{j,1}) \\ &= a_{j,1} \sum_{k \neq j} (1-a_{k,1}) + (1-a_{j,1}) \sum_{k \neq j} a_{k,1} \\ &= a_{j,1}(8+a_{j,1}) + (1-a_{j,1})^2 \\ &= M^2 + 8M + (1-M)^2 \\ &= 2M^2 + 6M + 1 \ge 2 \left( \frac{1}{100} + \frac{3}{10} \right) + 1 = \frac{81}{50}. \end{align*}$$
There are $9 \times 9 = 81$ summands in the above sum. Hence we can find $k,l$ such that
$$a_{j,1}a_{k,l} + a_{j,l}a_{k,1} \ge \frac{1}{50}$$