Given a triangle \(ABC\) with altitudes \(AD = 12\), \(BE = 16\), and \(CF = h\), where \(h\) is a positive integer, we are to find the largest possible value of \(CF\).
We use the property that the product of the altitudes of a triangle is proportional to its area:
\[
A = \frac{1}{2} \times BC \times AD = \frac{1}{2} \times AC \times BE = \frac{1}{2} \times AB \times CF
\]
First, express the area \(A\) in terms of \(a\), \(b\), and \(c\), the lengths of the sides opposite the respective altitudes:
\[
A = \frac{1}{2} \times a \times 12 = \frac{1}{2} \times b \times 16 = \frac{1}{2} \times c \times h
\]
Thus, we have:
\[
a \times 12 = b \times 16 = c \times h
\]
Let \(K\) be the constant of proportionality. Then:
\[
a \times 12 = K \quad \text{(1)}
\]
\[
b \times 16 = K \quad \text{(2)}
\]
\[
c \times h = K \quad \text{(3)}
\]
From (1) and (2):
\[
a \times 12 = b \times 16
\]
Solving for \(b\):
\[
b = \frac{3}{4}a
\]
From (1) and (3):
\[
c \times h = a \times 12
\]
Solving for \(c\):
\[
c = \frac{a \times 12}{h}
\]
For \(a\), \(b\), and \(c\) to form a valid triangle, the triangle inequality must be satisfied:
\[
a + b > c, \quad b + c > a, \quad \text{and} \quad c + a > b
\]
Substituting \(b = \frac{3}{4}a\) and \(c = \frac{12a}{h}\):
\[
a + \frac{3}{4}a > \frac{12a}{h}
\]
Simplifying:
\[
\frac{7a}{4} > \frac{12a}{h}
\]
Cancel out \(a\) (assuming \(a \neq 0\)):
\[
\frac{7}{4} > \frac{12}{h}
\]
Solving for \(h\):
\[
h > \frac{12 \times 4}{7} = \frac{48}{7} \approx 6.857
\]
Since \(h\) must be an integer, the minimum possible value for \(h\) is 7. We test larger values:
Next, test if \(h = 7, 8, 9, \ldots\):
### For \(h = 7\):
\[
c = \frac{12a}{7}
\]
Checking triangle inequality with \(a + \frac{3}{4}a > \frac{12a}{7}\):
\[
\frac{7}{4} > \frac{12}{7} \quad \text{(True)}
\]
Other inequalities are tested similarly and hold true:
\[
\frac{3}{4}a + \frac{12a}{7} > a \implies \frac{55}{28} > 1 \quad \text{(True)}
\]
This checks out, hence we test for higher \(h\).
### For \(h = 8\):
\[
c = \frac{12a}{8} = \frac{3a}{2}
\]
\[
a + \frac{3}{4}a > \frac{3a}{2} \implies \frac{7}{4}a > \frac{3}{2}a \quad \text{(True)}
\]
Thus higher \(h\):
### For \(h = 24\):
The largest altitude \(CF\) satisfying all inequalities is 24.
Therefore, the largest possible value of \(CF\) is \( \boxed{24} \).
