1. Assume the contrary: Suppose there does not exist an $n_0$ such that the flea can reach any interval $[n, n+1]$ with $n \geq n_0$. This means there are infinitely many intervals $[n, n+1]$ that the flea cannot jump into.

2. Define sequences: Let $(x_n)_{n \geq 0}$ be the infimum of these intervals. Since there are infinitely many such intervals, $x_n \to \infty$.

3. Greatest reachable point: Let $(y_n)_{n \geq 0}$ be the greatest number of the form $aX + bY$ (where $X, Y \in \mathbb{N}$) such that $y_n < x_n$. Clearly, $y_n \to \infty$.

4. **Representation of $y_n$**: We can write $y_n = a u_n + b v_n$ for some $u_n, v_n \in \mathbb{N}$. Since $y_n \to \infty$, at least one of $u_n$ or $v_n$ must tend to infinity. Without loss of generality, assume $v_n \to \infty$.

5. Density argument: Since $\frac{a}{b} \notin \mathbb{Q}$, by a form of Kronecker's theorem, the set $\{ au + bv \mid u \in \mathbb{N}, v \in \mathbb{Z} \}$ is dense in $\mathbb{R}$. Therefore, we can choose $u > 0$ and $v \in \mathbb{Z}$ such that $0 < au - vb < 1$.

6. **Choosing $N$**: Since $v_n \to \infty$, we can choose $N$ such that $v_N > v$. Hence, $y_N + (au - vb)$ is a position where the flea can jump.

7. Analyzing the jump:
- If $y_N + (au - vb) \geq x_N$, then $y_N + (au - vb) \in [x_N, x_N + 1]$, which contradicts our assumption that the flea cannot land in $[x_N, x_N + 1]$.
- If $y_N + (au - vb) < x_N$, then $y_N < y_N + (au - vb) < x_N$, which contradicts the fact that $y_N$ is the greatest number of the form $aX + bY$ less than $x_N$.

8. Conclusion: Both cases lead to a contradiction, so our initial assumption must be false. Therefore, there exists $n_0$ such that the flea can reach any interval $[n, n+1]$ with $n \geq n_0$.

$\blacksquare$