For $m \geqslant 2$ let us denote $s_{m}=\sum_{i+j=m} f(i, j)$. Since the remainder of $f(i, j)$ upon division by 2 is equal to $f(i, j)-2\left[\frac{f(i, j)}{2}\right]$, the number of odd numbers among the numbers $f(i, j)$ for $i, j \geqslant 0$ and $i+j=m$ is equal to
$$\begin{aligned} \sum_{i+j=m}\left(f(i, j)-2\left[\frac{f(i, j)}{2}\right]\right) & =s_{m}-\sum_{i+j=m}\left(\left[\frac{f(i-1, j+1)}{2}\right]+\left[\frac{f(i, j)}{2}\right]\right) \\ & =s_{m}-s_{m+1} \end{aligned}$$
It follows that the number of pairs $(i, j)$ for which $f(i, j)$ is odd and $i+j<m$ is equal to $s_{2}-s_{m}=n-s_{m}$.
It remains to show that $s_{m}=0$ for all sufficiently large $m$. It is clear that the sequence $s_{m}$ is nonnegative and nonincreasing, so there exist $N$ and $k$ such that $s_{m}=k$ for all $m \geqslant N$. This means that $f(i, j)$ is even whenever $i+j \geqslant m$. Suppose that $k>0$ and consider the smallest $i$ such that $f(i, m-i)>0$. By a simple induction we obtain $f(i, m+r-i)=\left[\frac{f(i, m-i)}{2^{r}}\right]$ for $r \geqslant 1$. However, if $2^{r} \leqslant f(i, m-i)<2^{r+1}$, this gives $f(i, m+r-i)=1$, contrary to the assumption.

Second solution (U. Dinić). Let us place $n$ tokens at the point $(1,1)$ in the coordinate plane. At each step, from each point $(i, j)$ we shall move the integer part of half of its tokens to the points $(i+1, j)$ and $(i, j+1)$. Note that, if some row or column is nonempty at some moment, it will remain nonempty forever. Thus no token can leave the square $[1, n] \times [1, n]$, so the game ends after a finite number of steps.
It is not hard to see that after $i+j-2$ steps there are exactly $f(i, j)$ tokens at the point $(i, j)$. Moreover, the number $f(i, j)$ is odd if after the next step exactly one token remains at the point $(i, j)$. Since in the final position there are exactly $n$ tokens, it follows that among the terms of the sequence $f(i, j)$ there are $n$ odd numbers.