A line with negative slope passing through the point $(18,8)$ intersects the $x$ and $y$ axes at $(a,0)$ and $(0,b)$, respectively.  What is the smallest possible value of $a+b$?
1. We start by writing the equation of the line in slope-intercept form: 
   \[
   y = -mx + b
   \]
   Since the line passes through the point \((18, 8)\), we substitute \(x = 18\) and \(y = 8\) into the equation:
   \[
   8 = -18m + b
   \]
   Solving for \(b\), we get:
   \[
   b = 18m + 8
   \]

2. The equation of the line is now:
   \[
   y = -mx + (18m + 8)
   \]

3. To find the \(x\)-intercept, set \(y = 0\):
   \[
   0 = -mx + (18m + 8)
   \]
   Solving for \(x\), we get:
   \[
   mx = 18m + 8 \implies x = \frac{18m + 8}{m} = 18 + \frac{8}{m}
   \]
   Thus, the \(x\)-intercept is \((a, 0)\) where \(a = 18 + \frac{8}{m}\).

4. To find the \(y\)-intercept, set \(x = 0\):
   \[
   y = 18m + 8
   \]
   Thus, the \(y\)-intercept is \((0, b)\) where \(b = 18m + 8\).

5. We need to minimize the sum \(a + b\):
   \[
   a + b = \left(18 + \frac{8}{m}\right) + (18m + 8) = 26 + \frac{8}{m} + 18m
   \]

6. Let \(m' = -m\). Then we need to minimize:
   \[
   26 + \frac{8}{m'} + 18m'
   \]

7. By the AM-GM inequality, for any positive numbers \(x\) and \(y\):
   \[
   \frac{x}{a} + ay \geq 2\sqrt{xy}
   \]
   Applying this to our expression, we set \(x = 8\) and \(y = 18\), and we get:
   \[
   \frac{8}{m'} + 18m' \geq 2\sqrt{8 \cdot 18} = 2\sqrt{144} = 24
   \]
   Equality holds when:
   \[
   \frac{8}{m'} = 18m' \implies 8 = 18m'^2 \implies m'^2 = \frac{8}{18} = \frac{4}{9} \implies m' = \frac{2}{3}
   \]

8. Substituting \(m' = \frac{2}{3}\) back into the expression for \(a + b\):
   \[
   26 + 24 = 50
   \]

The final answer is \(\boxed{50}\)