2. Let $n$ be a natural number. We define $f(0, j)=f(i, 0)=0, f(1,1)=n$ and

$$
f(i, j)=\left\lfloor\frac{f(i-1, j)}{2}\right\rfloor+\left\lfloor\frac{f(i, j-1)}{2}\right\rfloor
$$

for all natural numbers $i$ and $j, (i, j) \neq (1,1)$. How many ordered pairs of natural numbers $(i, j)$ are there for which $f(i, j)$ is an odd number? (Dusan Dukic)
2. The answer is $n$.

For $m \geqslant 2$ let $s_{m}=\sum_{i+j=m} f(i, j)$. Since the remainder of $f(i, j)$ when divided by 2 is $f(i, j)-2\left[\frac{f(i, j)}{2}\right]$, the number of odd numbers among the numbers $f(i, j)$ for $i, j \geqslant 0$ and $i+j=m$ is

$$
\begin{aligned}
\sum_{i+j=m}\left(f(i, j)-2\left[\frac{f(i, j)}{2}\right]\right) & =s_{m}-\sum_{i+j=m}\left(\left[\frac{f(i-1, j+1)}{2}\right]+\left[\frac{f(i, j)}{2}\right]\right) \\
& =s_{m}-s_{m+1}
\end{aligned}
$$

It follows that the number of pairs $(i, j)$ for which $f(i, j)$ is odd and $i+j>0$ and consider the smallest $i$ such that $f(i, m-i)>0$. By simple induction, we get $f(i, m+r-i)=\left[\frac{f(i, m-i)}{2^{r}}\right]$ for $r \geqslant 1$. However, if $2^{r} \leqslant f(i, m-i)<2^{r+1}$, then $f(i, m+r-i)=1$, contradicting the assumption.

Second solution (U. Dinić). Place $n$ tokens at the point $(1,1)$ in the coordinate plane. In each step, from each point $(i, j)$, we will move the integer part of half of its tokens to the points $(i+1, j)$ and $(i, j+1)$. Note that if a row or column is non-empty at some point, it will always remain non-empty. Thus, no token can leave the square $[1, n] \times[1, n]$, so the game ends in 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
there is one token left at the point $(i, j)$. Since there are exactly $n$ tokens in the final position, it follows that among the members of the sequence $f(i, j)$ there are $n$ odd numbers.