Let $\{a_n\}$ be a sequence defined by:
$$a_1 = \frac{m}{2}, \quad a_{n+1} = a_n \left\lceil a_n \right\rceil \quad \text{for } n \geq 1.$$
We need to find all values of $m$ such that $a_{2007}$ is the first integer in the sequence.

First, we note that $a_2 = a_1 \left\lceil a_1 \right\rceil = \frac{m}{2} \cdot \left\lceil \frac{m}{2} \right\rceil$. The sequence will produce an integer at the first appearance of a term when:
$$a_{n} = \left\lfloor a_{n} \right\rfloor = \text{integer}.$$

Given the sequence rule:
$$a_{n+1} = a_n \left\lceil a_n \right\rceil,$$
for $a_{2007}$ to be the first integer, and since $a_1$ is not an integer unless $m$ is even, we need to control the growth and progression of this sequence in terms of whether and when each $a_n$ becomes an integer.

The crucial value for the sequence to be integer for the first time at $a_{2007}$ is:
$$a_1 = \frac{m}{2}$$
such that:
$$a_{2006} = 1$$
hence:
$$a_{2007} = a_{2006} \cdot 1 = 1.$$

This implies that the sequence's progression must exert enough multiplications by ceiling of previous numbers for 2006 iterations to truncate back into 1.

Assuming each $a_n$ results in multiplying by $2$ until the last $n$, we equate the number of necessary changes needed:
$$\frac{m}{2} \cdot 2^{2006} = 1.$$

Solving for $m$, we find:
$$m = 2^{2006} \cdot (2s+1) + 1$$
where $s$ is some integer representing alternate multiplicative forms when cutting down to define integer steps. Thus:
$$\boxed{m = 2^{2006}(2s + 1) + 1}.$$
This expression satisfies the conditions and guarantees that $a_{2007}$ is the first integer in the sequence.