Assume we have $95$ boxes and $19$ balls distributed in these boxes in an arbitrary manner. We take $6$ new balls at a time and place them in $6$ of the boxes, one ball in each of the six. Can we, by repeating this process a suitable number of times, achieve a situation in which each of the $95$ boxes contains an equal number of balls?
1. **Initial Setup**: We start with 95 boxes and 19 balls distributed arbitrarily among these boxes. We are allowed to add 6 new balls at a time, placing one ball in each of 6 different boxes.

2. **Objective**: We need to determine if it is possible to achieve a situation where each of the 95 boxes contains an equal number of balls by repeating the process of adding 6 balls at a time.

3. **Total Balls Calculation**: 
   - Initially, there are 19 balls.
   - Each time we add 6 balls, the total number of balls increases by 6.
   - Let \( n \) be the number of times we add 6 balls. Then the total number of balls after \( n \) additions is \( 19 + 6n \).

4. **Equal Distribution Condition**: 
   - For all boxes to have an equal number of balls, the total number of balls must be divisible by 95.
   - Therefore, \( 19 + 6n \) must be divisible by 95.

5. **Divisibility Condition**:
   - We need to find \( n \) such that \( 19 + 6n \equiv 0 \pmod{95} \).
   - This simplifies to \( 6n \equiv -19 \pmod{95} \).

6. **Solving the Congruence**:
   - We need to find the multiplicative inverse of 6 modulo 95.
   - Using the Extended Euclidean Algorithm:
     \[
     95 = 15 \times 6 + 5 \\
     6 = 1 \times 5 + 1 \\
     5 = 5 \times 1 + 0
     \]
     Back-substituting:
     \[
     1 = 6 - 1 \times (95 - 15 \times 6) \\
     1 = 16 \times 6 - 95
     \]
     Thus, the multiplicative inverse of 6 modulo 95 is 16.

7. **Finding \( n \)**:
   - Multiply both sides of \( 6n \equiv -19 \pmod{95} \) by 16:
     \[
     n \equiv -19 \times 16 \pmod{95} \\
     n \equiv -304 \pmod{95} \\
     n \equiv 81 \pmod{95}
     \]
   - Therefore, \( n = 81 + 95k \) for any integer \( k \).

8. **Verification**:
   - For \( n = 81 \), the total number of balls is \( 19 + 6 \times 81 = 19 + 486 = 505 \).
   - Since \( 505 \div 95 = 5 \), each box will have exactly 5 balls.

Conclusion:
By repeating the process of adding 6 balls at a time 81 times, we can achieve a situation where each of the 95 boxes contains an equal number of balls.

The final answer is \( \boxed{ n = 81 + 95k } \) for any integer \( k \).