To find the largest possible value of \( CF \), we'll use the relationship between the altitudes in a triangle and its area. The area \( \Delta \) of a triangle can be expressed using any of its altitudes:

\[

\Delta = \frac{1}{2} \times \text{base} \times \text{height}

\]

For triangle \( ABC \), we have the following relationships:

- \( \Delta = \frac{1}{2} \times BC \times AD = \frac{1}{2} \times CA \times BE = \frac{1}{2} \times AB \times CF \)

Let the side lengths be \( a = BC \), \( b = CA \), and \( c = AB \). Then:

\[

\Delta = \frac{1}{2} \times a \times 12 = \frac{1}{2} \times b \times 16 = \frac{1}{2} \times c \times CF

\]

This simplifies to:

\[

\Delta = 6a = 8b = \frac{1}{2} c \times CF

\]

Equating these expressions:

\[

6a = 8b \quad \text{and} \quad 6a = \frac{1}{2} c \times CF

\]

### Step 1: Solve for \( a \) and \( b \)

From \( 6a = 8b \):

\[

\frac{a}{b} = \frac{8}{6} = \frac{4}{3}

\]

Thus, \( a = \frac{4}{3}b \).

### Step 2: Substitute into \( 6a = \frac{1}{2} c \times CF \)

Substitute \( a = \frac{4}{3}b \) into \( 6a = \frac{1}{2} c \times CF \):

\[

6 \times \frac{4}{3}b = \frac{1}{2} c \times CF

\]

Simplifying:

\[

8b = \frac{1}{2} c \times CF \quad \text{so} \quad 16b = c \times CF

\]

### Step 3: Find the Maximum Value of \( CF \)

We need to maximize \( CF \), which is a positive integer. Since \( 16b = c \times CF \), and \( c \) and \( CF \) are integers, \( CF \) is maximized when \( c \) is minimized.

Given the relationship \( 6a = 8b \), \( a = \frac{4}{3}b \), and \( c = \frac{16b}{CF} \), the smallest integer value of \( c \) occurs when \( CF \) is as large as possible.

If \( CF = 16 \), then:

\[

16b = c \times 16 \quad \text{so} \quad c = b

\]

Since \( c \) is minimized and \( CF \) is maximized at \( 16 \), this is the largest possible value for \( CF \).

Thus, the largest possible value of \( CF \) is \( \boxed{24} \).