1. Restate the Problem in Grid Terms:
- Alice and Bob are playing a game on a $4 \times 9$ grid.
- Alice selects 18 cells to color black.
- Bob attempts to pair as many pairs of squares which are differently colored and lie in the same row or column as possible.
- We need to find the maximum number of pairs Bob can guarantee, regardless of Alice's actions.

2. Upper Bound on Bob's Points:
- Alice can prevent Bob from getting more than 15 pairs.
- If Alice colors the upper-left $3 \times 6$ grid black, Bob cannot pair the three squares in the lower-right $1 \times 3$ and hence can only get at most 15 pairs.

3. Lower Bound on Bob's Points:
- Let $x_i$ be the number of columns with exactly $i$ black squares, for $0 \le i \le 4$.
- We have the equation $x_0 + x_1 + x_2 + x_3 + x_4 = 9$.

4. Pairing Strategy:
- Bob can pair up all the squares in a column with 2 black squares.
- Bob can always pair up all the squares in any two columns with a total of 4 black squares.

5. Balancing the Columns:
- If $x_0 = x_4$, then $x_1 = x_3$ and we are done.
- This implies that Bob can leave at most 6 squares unpaired.

6. Conclusion:
- Since Bob can always ensure that at most 6 squares are unpaired, he can guarantee at least $36 - 6 = 30$ pairs.
- However, since each pair consists of two cards, the number of points Bob can guarantee is $\frac{30}{2} = 15$.

$\blacksquare$

The final answer is $\boxed{15}$