1. Define the problem and notation:
We need to determine the maximal number of consecutive interesting integers. An integer $k$ is interesting if the product of the first $k$ primes is divisible by $k$. Let $p_n$ denote the $n$-th smallest prime number. For example, $p_1 = 2$, $p_2 = 3$, $p_3 = 5$, and so on.

2. **Check small values of $k$:**
- For $k = 1$, the product of the first $1$ prime is $p_1 = 2$, and $2$ is divisible by $1$. Hence, $1$ is interesting.
- For $k = 2$, the product of the first $2$ primes is $p_1 \cdot p_2 = 2 \cdot 3 = 6$, and $6$ is divisible by $2$. Hence, $2$ is interesting.
- For $k = 3$, the product of the first $3$ primes is $p_1 \cdot p_2 \cdot p_3 = 2 \cdot 3 \cdot 5 = 30$, and $30$ is divisible by $3$. Hence, $3$ is interesting.
- For $k = 4$, the product of the first $4$ primes is $p_1 \cdot p_2 \cdot p_3 \cdot p_4 = 2 \cdot 3 \cdot 5 \cdot 7 = 210$, and $210$ is not divisible by $4$. Hence, $4$ is not interesting.

3. Generalize the observation:
- For $k = 5$, the product of the first $5$ primes is $p_1 \cdot p_2 \cdot p_3 \cdot p_4 \cdot p_5 = 2 \cdot 3 \cdot 5 \cdot 7 \cdot 11 = 2310$, and $2310$ is divisible by $5$. Hence, $5$ is interesting.
- For $k = 6$, the product of the first $6$ primes is $p_1 \cdot p_2 \cdot p_3 \cdot p_4 \cdot p_5 \cdot p_6 = 2 \cdot 3 \cdot 5 \cdot 7 \cdot 11 \cdot 13 = 30030$, and $30030$ is divisible by $6$. Hence, $6$ is interesting.
- For $k = 7$, the product of the first $7$ primes is $p_1 \cdot p_2 \cdot p_3 \cdot p_4 \cdot p_5 \cdot p_6 \cdot p_7 = 2 \cdot 3 \cdot 5 \cdot 7 \cdot 11 \cdot 13 \cdot 17 = 510510$, and $510510$ is divisible by $7$. Hence, $7$ is interesting.
- For $k = 8$, the product of the first $8$ primes is $p_1 \cdot p_2 \cdot p_3 \cdot p_4 \cdot p_5 \cdot p_6 \cdot p_7 \cdot p_8 = 2 \cdot 3 \cdot 5 \cdot 7 \cdot 11 \cdot 13 \cdot 17 \cdot 19 = 9699690$, and $9699690$ is not divisible by $8$. Hence, $8$ is not interesting.

4. Conclusion:
From the above steps, we observe that the maximal number of consecutive interesting integers is $7$.

The final answer is $\boxed{ 7 }$.