CroatianCompetitions

Let us consider the binomial coefficients $\binom{2n+1}{k}$ for $1 \leq k \leq n$.

Recall that $\binom{2n+1}{k}$ is odd if and only if, in the binary representation, every bit of $k$ is less than or equal to the corresponding bit of $2n+1$ (Lucas' theorem).

But we can use a parity argument:

The sum of all binomial coefficients for a given $m$ is $2^m$:
$$\sum_{k=0}^{2n+1} \binom{2n+1}{k} = 2^{2n+1}.$$

But we are interested in the parity of $\binom{2n+1}{k}$ for $1 \leq k \leq n$.

Note that $\binom{2n+1}{k} = \binom{2n+1}{2n+1-k}$, so the sequence is symmetric about $k = n+\frac{1}{2}$.

For $k = 0$ and $k = 2n+1$, $\binom{2n+1}{0} = \binom{2n+1}{2n+1} = 1$ (odd).

Let $S$ be the set $\{\binom{2n+1}{1}, \binom{2n+1}{2}, \dots, \binom{2n+1}{n}\}$.

Let $T$ be the set $\{\binom{2n+1}{n+1}, \dots, \binom{2n+1}{2n}\}$.

Since $\binom{2n+1}{k} = \binom{2n+1}{2n+1-k}$, the set $S$ and $T$ have the same number of odd elements.

Now, the total number of odd binomial coefficients in the row $2n+1$ is a power of $2$ (Lucas' theorem):

Let $2n+1$ in binary have $d$ ones, then the number of odd binomial coefficients is $2^d$.

But $\binom{2n+1}{0}$ and $\binom{2n+1}{2n+1}$ are both $1$ (odd), so the number of odd binomial coefficients among $k=1$ to $2n$ is $2^d - 2$.

Since $S$ and $T$ are symmetric, the number of odd binomial coefficients in $S$ is $\frac{2^d - 2}{2} = 2^{d-1} - 1$.

But $2^{d-1} - 1$ is odd for $d \geq 1$.

Therefore, there is an odd number of odd numbers among $\binom{2n+1}{1}, \dots, \binom{2n+1}{n}$.