Solution:

We need to find the largest $k$ such that $2^{k}$ divides $27!$.

The exponent of $2$ in $27!$ is given by:
$$\left\lfloor \frac{27}{2} \right\rfloor + \left\lfloor \frac{27}{4} \right\rfloor + \left\lfloor \frac{27}{8} \right\rfloor + \left\lfloor \frac{27}{16} \right\rfloor + \left\lfloor \frac{27}{32} \right\rfloor$$
Calculate each term:

$\left\lfloor \frac{27}{2} \right\rfloor = 13$

$\left\lfloor \frac{27}{4} \right\rfloor = 6$

$\left\lfloor \frac{27}{8} \right\rfloor = 3$

$\left\lfloor \frac{27}{16} \right\rfloor = 1$

$\left\lfloor \frac{27}{32} \right\rfloor = 0$

Sum:
$$13 + 6 + 3 + 1 + 0 = 23$$

Therefore, the largest positive integer $k$ such that $27!$ is divisible by $2^{k}$ is $k = 23$.