A grasshopper rests on the point $(1,1)$ on the plane. Denote by $O,$ the origin of coordinates. From that point, it jumps to a certain lattice point under the condition that, if it jumps from a point $A$ to $B,$ then the area of $\triangle AOB$ is equal to $\frac 12.$
$(a)$ Find all the positive integral poijnts $(m,n)$ which can be covered by the grasshopper after a finite number of steps, starting from $(1,1).$
$(b)$ If a point $(m,n)$ satisfies the above condition, then show that there exists a certain path for the grasshopper to reach $(m,n)$ from $(1,1)$ such that the number of jumps does not exceed $|m-n|.$
1. **Understanding the Problem:**
   - The grasshopper starts at point \((1,1)\).
   - It can jump to another lattice point \((m,n)\) such that the area of \(\triangle AOB\) is \(\frac{1}{2}\), where \(A\) is the starting point and \(B\) is the destination point.
   - We need to find all positive integral points \((m,n)\) that the grasshopper can reach and show that the number of jumps does not exceed \(|m-n|\).

2. **Condition for the Area:**
   - The area of \(\triangle AOB\) is given by:
     \[
     \text{Area} = \frac{1}{2} \left| a_1 b_2 - a_2 b_1 \right|
     \]
   - For the area to be \(\frac{1}{2}\), we need:
     \[
     \left| a_1 b_2 - a_2 b_1 \right| = 1
     \]

3. **Finding Reachable Points:**
   - We start from \((1,1)\) and need to find all \((m,n)\) such that \(\gcd(m,n) = 1\).
   - If \(\gcd(m,n) > 1\), then \((m,n)\) is not reachable because the determinant condition \(\left| a_1 b_2 - a_2 b_1 \right| = 1\) cannot be satisfied.

4. **Induction Proof for Reachable Points:**
   - **Base Case:** For \(m+n = 2\), the only point is \((1,1)\), which is trivially reachable.
   - **Inductive Step:** Assume the statement is true for all pairs \((m',n')\) such that \(m' + n' < m + n\).
     - We need to show it for \((m,n)\) with \(\gcd(m,n) = 1\).
     - There exists a positive integer \(p\) such that \(1 \leq p \leq n\) and \(mp \equiv 1 \pmod{n}\).
     - This implies \(m(n-p) \equiv -1 \pmod{n}\).
     - One of \(p\) or \(n-p\) is less than \(\frac{n}{2}\). Let \(q\) be the smaller of \(p\) or \(n-p\).
     - Then \(mq \equiv \pm 1 \pmod{n}\), so there exists an integer \(r\) such that \(mq - nr = \pm 1\).
     - This implies \(\gcd(q,r) = 1\) and \(q + r < m + n\).
     - By the induction hypothesis, \((q,r)\) is reachable.
     - Since \((q,r)\) and \((m,n)\) form a triangle with area \(\frac{1}{2}\), \((m,n)\) is also reachable.

5. **Number of Jumps:**
   - For a point \((m,n)\) with \(m \geq n\), we construct a smaller point \((q,r)\) such that \(mq - nr = \pm 1\).
   - This gives \((m-n)q - n(r-q) = \pm 1\).
   - Since \(q < \frac{n}{2}\), we have \(r - q < m - n\).
   - By induction, \((r,q)\) can be reached in \(r - q\) jumps.
   - Therefore, \((m,n)\) can be reached in \(r - q + 1 \leq m - n\) jumps.

The final answer is \( \boxed{ (m,n) } \) such that \(\gcd(m,n) = 1\).