Ketten play the following game according to the system below:

$$
\begin{aligned}
& a_{1} x_{1}+\ldots+a_{n} x_{n}=c \\
& b_{1} x_{1}+\ldots+b_{n} x_{n}=d
\end{aligned}
$$

They take turns writing arbitrary real numbers in place of the $a_{i}, b_{i}, x_{i}(i=1,2, \ldots, n)$ and $c, d$ letters. However, they can only write 0 in place of a letter if there is a row in which this letter appears, and all other letters in that row have already been assigned values. The player who starts wants to achieve that both equalities are valid.

For which $n$ can they achieve this, no matter what the other does?
If a player completes a certain equation, that is, the player assigns a value to the last free variable in that equation, and the coefficient of this variable is not zero, then the equation is solvable for this last variable as the unknown. Choosing this unique solution ensures that the equation holds, while any other value would disrupt it.

Any player can ensure that the zero value does not appear before completing any equation. If the player in question keeps choosing from the $x_i$ variables until there are no free ones left, then in at most $n$ steps, each $x_i$ will be chosen in sequence, and in the meantime, the opponent will also have taken at most $n$ steps, meaning that in total, at most $2n$ variables have been chosen. Since each equation contains $2n+1$ variables, according to the problem's conditions, the players cannot have chosen the zero value yet.

The total number of variables is $3n+2$, which is even or odd depending on whether $n$ is even or odd. If $n$ is even, the second player finishes the game, meaning they will be the last to choose in at least one of the equations, and thus, according to the above, can ensure that the first player does not win. Let $n$ be odd. Now the first player plays according to the above method, ensuring that in the first $2k$ steps $(k \leq n)$, the $x$ variables run out, and then aims to complete both equations. The second player, with their $2k$-th and every subsequent move, assigns values to an even number of variables, so the number of free variables in one equation remains even, and in the other, it remains odd (the free variables differ between the equations because the common variables are the $x$s, which have run out). The first player now always chooses in the equation where the number of free variables is odd. This way, they complete both equations and win the game. Therefore, the first player can ensure a win if and only if $n$ is odd.

Balassa András (Budapest, Fazekas M. Gyak. Gimn., II. o. t.)