Let's analyze the information provided about Pedro and Ana to solve the problem and find out which numbers Julian removed.

### Pedro's Drawn Numbers

We are told that Pedro picks three consecutive numbers whose product is 5 times their sum. Let $a$, $a+1$, and $a+2$ be the consecutive numbers drawn by Pedro. Therefore, we have:

- Product:
$$a(a+1)(a+2)$$

- Sum:
$$a + (a+1) + (a+2) = 3a + 3 = 3(a + 1)$$

According to the problem, the product is 5 times the sum:
$$a(a+1)(a+2) = 5 \times 3(a + 1)$$
Simplifying gives:
$$a(a+1)(a+2) = 15(a + 1)$$

Divide both sides by $a+1$ (assuming $a \neq -1$):
$$a(a+2) = 15$$
$$a^2 + 2a - 15 = 0$$

Solving the quadratic equation:
$$(a-3)(a+5) = 0$$
Thus, $a = 3$ or $a = -5$. The value $a = -5$ is not possible as it does not fit the range, therefore Pedro's numbers are:
$3, 4, 5$.

### Ana's Drawn Numbers

Ana mentions she picked only non-prime numbers, where two of them are consecutive and the product is 4 times the sum. The non-prime numbers in the range 1 to 9 are $1, 4, 6, 8, 9$.

Let $b, c, d$ be Ana's numbers, such that two are consecutive. We need to test combinations fitting the condition that:
$$bcd = 4(b+c+d)$$

By testing combinations, let's choose $4, 6$ as Ana's consecutive non-prime numbers:

Combining with 8:
- Product:
$$4 \times 6 \times 8 = 192$$

- Sum:
$$4 + 6 + 8 = 18$$

Verification:
$$192 = 4 \times 18 = 72$$

It doesn’t satisfy.

Now try combining $4, 6, 9$:
- Product:
$$4 \times 6 \times 9 = 216$$

- Sum:
$$4 + 6 + 9 = 19$$

Verification:
$$216 = 4 \times 19$$

Thus, Ana’s numbers are $4, 6, 9$.

### Julian's Drawn Numbers

Since Pedro has picked $3, 4, 5$ and Ana has picked $4, 6, 9$, the remaining numbers for Julian, drawn from 1 to 9, that have not been picked are:
- Remaining numbers: $1, 2, 6, 7, 8$.

Since Ana already took 6, 9, and needs two consecutive:
- The consecutive pair 6 and 7 fits for Julian

Let's verify if Julian's numbers can reasonably be drawn:
Therefore, Julian’s numbers are $\boxed{2, 6, 7}$.