Solution:
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 $k<n$ in order for the frog to be able to make its first jump. As the frog jumps to the right only if it is in a number $a \leqslant 0$, and the largest jump has length $n$, it is impossible to reach numbers greater than $n$. On the other hand, suppose the frog is in a number $a>0$, 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
$$k-n+1 \leqslant i \leqslant 0 \quad \text{or} \quad k+1 \leqslant i \leqslant n.$$
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 $2k+1 \leqslant a \leqslant n$, therefore $2k+1 \leqslant n$. 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 $2k-n+1 \leqslant a+k \leqslant 0$, so in this case we have $2k+1 \leqslant n$ 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 upper bound 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 $n=2k+1$. We claim that when the frog performs the jumps of length $1, \ldots, 2k+1$ 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 $n=2k+1$ 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 $k+2, k+3, \ldots, 2k+1$, 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 $(k+1)-(k+(i+1))=-i<1$, after which it lands in $(-i)+i=0$. We see that, starting in $0$, the frog lands in $k+1$ after the pair of jumps, while starting in $k+1$ the frog lands in $0$, while in both cases the jumps do not touch $1,2, \ldots, k$. This proves the correctness of its series of jumps. As the frog (after its first jump) alternates between $k+1$ and $0$ exactly $k$ times, for odd $k$ it will end up in $0$, while for even $k$ it will end up in $k+1$.

Suppose $n$ is even, then $\frac{n-1}{2}$ is not an integer and we have $k=\frac{n-1}{2}-\frac{1}{2}=\frac{n-2}{2}$, so $n=2k+2$. Let the frog firstly perform the same series of jumps as in the previous case; they still do not touch $1,2, \ldots, k$. Now let the frog make a final extra jump of length $2k+2$. It will land in $0+(2k+2)=2k+2>k$ if $k$ is odd, or in $(k+1)-(2k+2)=-k-1<1$ 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 $\left\lfloor\frac{n-1}{2}\right\rfloor$.