Solution:

The initial force \( F \) between two charges \( q_1 \) and \( q_2 \) is given by:

\[
F = k \frac{q_1 q_2}{r^2} = 100 \, \text{N}
\]

After changing the charges:
- \( q_1 \) increases by 10%, so \( q_1' = 1.1 q_1 \).
- \( q_2 \) decreases by 10%, so \( q_2' = 0.9 q_2 \).

The new force \( F' \) is:

\[
F' = k \frac{q_1' q_2'}{r^2} = k \frac{(1.1 q_1)(0.9 q_2)}{r^2} = (1.1 \times 0.9) F
\]

Calculating \( 1.1 \times 0.9 = 0.99 \), so:

\[
F' = 0.99 \times 100 = 99 \, \text{N}
\]