The harmonic table is a triangular array:

$1$

$\frac 12 \qquad \frac 12$

$\frac 13 \qquad \frac 16 \qquad \frac 13$

$\frac 14 \qquad \frac 1{12} \qquad \frac 1{12} \qquad \frac 14$

Where $a_{n,1} = \frac 1n$ and $a_{n,k+1} = a_{n-1,k} - a_{n,k}$ for $1 \leq k \leq n-1.$ Find the harmonic mean of the $1985^{th}$ row.
1. **Identify the elements of the \( n \)-th row:**
   The elements of the \( n \)-th row are given by:
   \[
   a_{n,1} = \frac{1}{n}, \quad a_{n,k+1} = a_{n-1,k} - a_{n,k} \quad \text{for} \quad 1 \leq k \leq n-1
   \]
   We need to show that these elements are:
   \[
   \frac{1}{n\binom{n-1}{0}}, \frac{1}{n\binom{n-1}{1}}, \cdots, \frac{1}{n\binom{n-1}{n-1}}
   \]

2. **Base cases verification:**
   For \( n = 1 \):
   \[
   a_{1,1} = \frac{1}{1} = 1
   \]
   For \( n = 2 \):
   \[
   a_{2,1} = \frac{1}{2}, \quad a_{2,2} = a_{1,1} - a_{2,1} = 1 - \frac{1}{2} = \frac{1}{2}
   \]
   For \( n = 3 \):
   \[
   a_{3,1} = \frac{1}{3}, \quad a_{3,2} = a_{2,1} - a_{3,1} = \frac{1}{2} - \frac{1}{3} = \frac{1}{6}, \quad a_{3,3} = a_{2,2} - a_{3,2} = \frac{1}{2} - \frac{1}{6} = \frac{1}{3}
   \]
   For \( n = 4 \):
   \[
   a_{4,1} = \frac{1}{4}, \quad a_{4,2} = a_{3,1} - a_{4,1} = \frac{1}{3} - \frac{1}{4} = \frac{1}{12}, \quad a_{4,3} = a_{3,2} - a_{4,2} = \frac{1}{6} - \frac{1}{12} = \frac{1}{12}, \quad a_{4,4} = a_{3,3} - a_{4,3} = \frac{1}{3} - \frac{1}{12} = \frac{1}{4}
   \]

3. **Inductive step:**
   Assume the elements of the \( n \)-th row are:
   \[
   \frac{1}{n\binom{n-1}{0}}, \frac{1}{n\binom{n-1}{1}}, \cdots, \frac{1}{n\binom{n-1}{n-1}}
   \]
   We need to show that the elements of the \( (n+1) \)-th row are:
   \[
   \frac{1}{(n+1)\binom{n}{0}}, \frac{1}{(n+1)\binom{n}{1}}, \cdots, \frac{1}{(n+1)\binom{n}{n}}
   \]
   Using the recurrence relation:
   \[
   a_{n+1,k+1} = a_{n,k} - a_{n+1,k}
   \]
   We have:
   \[
   \frac{1}{n\binom{n-1}{k-1}} - \frac{1}{(n+1)\binom{n}{k-1}} = \frac{1}{(n+1)\binom{n}{k}}
   \]
   This confirms the inductive step.

4. **Harmonic mean calculation:**
   The harmonic mean of the elements of the \( n \)-th row is:
   \[
   \frac{n}{\sum_{k=0}^{n-1} \frac{1}{n\binom{n-1}{k}}} = \frac{n}{\frac{1}{n} \sum_{k=0}^{n-1} \binom{n-1}{k}} = \frac{n}{\frac{1}{n} \cdot 2^{n-1}} = \frac{n^2}{2^{n-1}}
   \]
   Simplifying, we get:
   \[
   \frac{1}{2^{n-1}}
   \]

5. **Harmonic mean of the 1985th row:**
   For \( n = 1985 \):
   \[
   \frac{1}{2^{1984}}
   \]

The final answer is \( \boxed{ \frac{1}{2^{1984}} } \)