. Let \( n \geqslant 3 \) be an integer. A frog is to jump along the real axis, starting at the point 0 and making \( n \) jumps: one of length 1, one of length 2, \(\ldots\), one of length \( n \). It may perform these \( n \) jumps in any order. If at some point the frog is sitting on a number \( a \leqslant 0 \), its next jump must be to the right (towards the positive numbers). If at some point the frog is sitting on a number \( a > 0 \), its next jump must be to the left (towards the negative numbers). Find the largest positive integer \( k \) for which the frog can perform its jumps in such an order that it never lands on any of the numbers \( 1, 2, \ldots, k \).
We claim that the largest positive integer $k$ with the given property is $\left\lfloor\frac{n-1}{2}\right\rfloor$, where $\lfloor x\rfloor$ is by definition the largest integer not exceeding $x$.

Consider a sequence of $n$ jumps of length $1,2, \ldots n$ such that the frog never lands on any of the numbers $1,2, \ldots, k$, where $k \geqslant 1$. Note that we must have $k0$, then it must even be in a number $a \geqslant k+1$, since it is not allowed to hit the numbers $1,2, \ldots, k$. So the frog jumps to the left only if it is in a number $a \geqslant k+1$, and therefore it is impossible to reach numbers less than $(k+1)-n=k-n+1$. This means the frog only possibly lands on the numbers $i$ satisfying

$$
k-n+1 \leqslant i \leqslant 0 \quad \text { or } \quad k+1 \leqslant i \leqslant n \text {. }
$$

When performing a jump of length $k$, the frog has to remain at either side of the numbers $1,2, \ldots, k$. Indeed, jumping over $1,2, \ldots, k$ requires a jump of at least length $k+1$. In case it starts at a number $a>0$ (in fact $k+1 \leqslant a \leqslant n$ ), it lands in $a-k$ and we must also have $a-k \geqslant k+1$. So $2 k+1 \leqslant a \leqslant n$, therefore $2 k+1 \leqslant n$. In case it starts at a number $a \leqslant 0$ (in fact $k-n+1 \leqslant a \leqslant 0$ ), it lands in $a+k$ and we must also have $a+k \leqslant 0$. Adding $k$ to both sides of $k-n+1 \leqslant a$, we obtain $2 k-n+1 \leqslant a+k \leqslant 0$, so in this case we have $2 k+1 \leqslant n$ as well. We conclude that $k \leqslant \frac{n-1}{2}$. Since $k$ is integer, we even have $k \leqslant\left\lfloor\frac{n-1}{2}\right\rfloor$.

Next we prove that this upperbound is sharp: for $k=\left\lfloor\frac{n-1}{2}\right\rfloor$ the frog really can perform its jumps in such an order that it never lands on any of the numbers $1,2, \ldots, k$.

Suppose $n$ is odd, then $\frac{n-1}{2}$ is an integer and we have $k=\frac{n-1}{2}$, so $n=2 k+1$. We claim that when the frog performs the jumps of length $1, \ldots, 2 k+1$ in the following order, it does never land on $1,2, \ldots, k$ : it starts with a jump of length $k+1$, then it performs two jumps, one of length $k+2$ followed by one of length 1 , next two jumps of length $k+3$ and $2, \ldots$, next two jumps of length $k+(i+1)$ and $i, \ldots$, and finally two jumps of length $k+(k+1)$ and $k$. In this order of the jumps every length between 1 and $n=2 k+1$ does occur: it performs a pair of jumps for $1 \leqslant i \leqslant k$, which are the jumps of length $1,2, \ldots$, $k$ and the jumps of length $k+2, k+3, \ldots, 2 k+1$, and it starts with the jump of length $k+1$.

We now prove the correctness of this jumping scheme. After the first jump the frog lands in $k+1>k$. Now suppose the frog is in 0 or $k+1$ and is about to perform the pair of jumps of length $k+(i+1)$ and $i$. Starting from 0 , it lands in $k+(i+1)>k$, after which it lands in $(k+i+1)-i=k+1>k$. If on the contrary it starts in $k+1$, it lands in $(k+1)-(k+(i+1))=-ik$ if $k$ is odd, or in $(k+1)-(2 k+2)=-k-1<1$ if $k$ is even, and its series of jumps is correct again.

We conclude that the largest positive integer $k$ with the given property is $\left\lfloor\frac{n-1}{2}\right\rfloor$.