Argentine National Olympiad, 2015

Yes, Roberto is right. To prove this, we first show that the least $n$ for which 1226 triminos $1\times 3$ can be placed on an $n \times n$ square without common points (including vertices) is $n=99$. Next, take a $99\times 99$ square. In its first row one can place 25 triminos without point in common; they leave uncovered the 24 cells whose position numbers are divisible by 4. Do the same with all odd-numbered rows 1, 3, 5, ..., 99, of which there are 50. This arrangement of $25 \cdot 50 = 1250$ triminos satisfies the conditions and, clearly, any greater grid square can accommodate 1250 trimino too.

So let $Q$ be an $n \times n$ square containing 1226 triminos $1\times 3$ without common points (including vertices). Extend $Q$ to an $(n+1) \times (n+1)$ square $Q'$ by adjoining an additional bottom row and additional leftmost column, with $n+1$ cells in each of them. To each trimino $T$ add 5 cells to the left and under it so that $T$ and these 5 cells form a rectangle $R(T)$ with dimensions $2\times 4$. The definition of $R(T)$ is illustrated in the figures, for horizontal and vertical triminos.

The $2\times 4$ rectangles defined in this way may stick out of $Q$ but are contained in the bigger square $Q'$.
The point of the construction is that the rectangles $R(T)$ do not overlap, i.e., no two of them have cells in common. Indeed, let $T$ and $T'$ be arbitrary triminos. By symmetry assume that $T$ is vertical. Enclose it in a $3\times 5$ rectangle as shown in the figure.

Note that by hypothesis this rectangle $R^*$ contains no cell of the trimino $T'$. The cells marked with $\bullet$ form the rectangle $R(T)$ together with $T$. Denote by $C$ the top right cell in $T'$; it is also the top right cell of $R(T')$.
Suppose that $C$ is under line $a$. Then so is the entire rectangle $R(T')$ by its definition, hence $R(T')$ does not overlap with $R(T)$. The same follows if $C$ is to the left of line $c$. Suppose that $C$ is in region I, i.e., between the lines $c, d$ and above line $b$. Then the entire $T'$ is above $b$. This is clear if $T'$ is horizontal. And if $T'$ is vertical then having cells under $b$ mean having cells in $R^*$ which is forbidden. So indeed the whole of $T'$ is above $b$, implying that $R(T')$ is above line $b'$. Then $R(T')$ does not overlap $R(T)$ again. The case where $C$ is in region II is analogous; here $T'$ is to the right of line $d$ and $R(T')$ is to the right of line $d'$. Finally the case of $C$ in region III is obvious.

In summary, we obtain 1226 non-overlapping rectangles $2 \times 4$ in the $(n+1) \times (n+1)$ square $Q'$. By area considerations then $(n+1)^2 \ge 8 \cdot 1226 = 9808$ and so $n+1 \ge \sqrt{9808} = 99.035...$. It follows that $n \ge 99$, as stated, so the starting discussion completes the proof.