Consider the Harmonic Table 
\[\begin{array}{c@{\hspace{15pt}}c@{\hspace{15pt}}c@{\hspace{15pt}}c@{\hspace{15pt}}c@{\hspace{15pt}}c@{\hspace{15pt}}c}&&&1&&&\\&&\tfrac12&&\tfrac12&&\\&\tfrac13&&\tfrac16&&\tfrac13&\\\tfrac14&&\tfrac1{12}&&\tfrac1{12}&&\tfrac14\\&&&\vdots&&&\end{array}\] where $a_{n,1}=1/n$ and \[a_{n,k+1}=a_{n-1,k}-a_{n,k}.\] Find the remainder when the sum of the reciprocals of the $2007$ terms on the $2007^\text{th}$ row gets divided by $2008$.
1. **Understanding the Harmonic Table**:
   The given table is defined by:
   \[
   a_{n,1} = \frac{1}{n}
   \]
   and
   \[
   a_{n,k+1} = a_{n-1,k} - a_{n,k}.
   \]
   We need to find the sum of the reciprocals of the 2007 terms in the 2007th row and then find the remainder when this sum is divided by 2008.

2. **Pattern Recognition**:
   Let's first try to understand the pattern in the table. We start by computing the first few rows:
   - 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 \( n = 3 \):
     \[
     a_{3,1} = \frac{1}{3}, \quad a_{3,2} = a_{2,1} - a_{3,1} = \frac{1}{2} - \frac{1}{3} = \frac{1}{6}, \quad a_{3,3} = a_{2,2} - a_{3,2} = \frac{1}{2} - \frac{1}{6} = \frac{1}{3}
     \]

3. **General Pattern**:
   We observe that the terms in the \( n \)-th row are of the form:
   \[
   a_{n,k} = \frac{1}{n \binom{n}{k}}
   \]
   This can be verified by induction or by observing the pattern in the table.

4. **Sum of Reciprocals**:
   The sum of the reciprocals of the 2007 terms in the 2007th row is:
   \[
   \sum_{k=1}^{2007} a_{2007,k} = \sum_{k=1}^{2007} \frac{1}{2007 \binom{2007}{k}}
   \]
   Using the identity:
   \[
   \sum_{k=0}^{n} \binom{n}{k} = 2^n
   \]
   We get:
   \[
   \sum_{k=1}^{2007} \frac{1}{2007 \binom{2007}{k}} = \frac{1}{2007} \sum_{k=1}^{2007} \frac{1}{\binom{2007}{k}} = \frac{1}{2007} \left( 2^{2007} - 1 \right)
   \]

5. **Finding the Remainder**:
   We need to find the remainder when:
   \[
   \frac{2^{2007} - 1}{2007}
   \]
   is divided by 2008. Since 2007 and 2008 are coprime, we can use Fermat's Little Theorem:
   \[
   2^{2007} \equiv 2 \pmod{2008}
   \]
   Therefore:
   \[
   2^{2007} - 1 \equiv 2 - 1 \equiv 1 \pmod{2008}
   \]
   Hence:
   \[
   \frac{2^{2007} - 1}{2007} \equiv \frac{1}{2007} \pmod{2008}
   \]
   Since 2007 and 2008 are coprime, the remainder is 1.

The final answer is \(\boxed{1}\).