Let $ \mathbb{N}$ denote the set of all positive integers. Prove that there exists a unique function $ f: \mathbb{N} \mapsto \mathbb{N}$ satisfying
\[ f(m \plus{} f(n)) \equal{} n \plus{} f(m \plus{} 95)
\]
for all $ m$ and $ n$ in $ \mathbb{N}.$ What is the value of $ \sum^{19}_{k \equal{} 1} f(k)?$
To prove that there exists a unique function \( f: \mathbb{N} \to \mathbb{N} \) satisfying the functional equation
\[ f(m + f(n)) = n + f(m + 95) \]
for all \( m \) and \( n \) in \( \mathbb{N} \), and to find the value of \( \sum_{k=1}^{19} f(k) \), we proceed as follows:

1. **Substitute and Simplify:**
   Let us denote the given functional equation as \( P(m, n) \):
   \[ P(m, n): f(m + f(n)) = n + f(m + 95) \]

2. **Initial Substitution:**
   Substitute \( m = 1 \) into \( P(m, n) \):
   \[ f(1 + f(n)) = n + f(1 + 95) \]
   Let \( c = f(1 + 95) \). Then,
   \[ f(1 + f(n)) = n + c \]

3. **Exploring the Form of \( f \):**
   Substitute \( m = 1 \) and \( n = 1 \) into \( P(m, n) \):
   \[ f(1 + f(1)) = 1 + f(1 + 95) \]
   \[ f(1 + f(1)) = 1 + c \]

4. **General Form:**
   Assume \( f(n) = n + k \) for some constant \( k \). Substitute this into the original equation:
   \[ f(m + f(n)) = f(m + n + k) = m + n + k + k = m + n + 2k \]
   \[ n + f(m + 95) = n + m + 95 + k \]
   Equating both sides, we get:
   \[ m + n + 2k = n + m + 95 + k \]
   \[ 2k = 95 + k \]
   \[ k = 95 \]

5. **Unique Function:**
   Thus, the function \( f(n) = n + 95 \) satisfies the given functional equation. To check uniqueness, assume there is another function \( g \) that satisfies the same equation. By similar steps, we would find that \( g(n) = n + 95 \) must hold, proving the uniqueness.

6. **Summation:**
   Now, we need to find the value of \( \sum_{k=1}^{19} f(k) \):
   \[ f(k) = k + 95 \]
   \[ \sum_{k=1}^{19} f(k) = \sum_{k=1}^{19} (k + 95) \]
   \[ = \sum_{k=1}^{19} k + \sum_{k=1}^{19} 95 \]
   \[ = \frac{19 \cdot 20}{2} + 19 \cdot 95 \]
   \[ = 190 + 1805 \]
   \[ = 1995 \]

The final answer is \( \boxed{1995} \)