4. Fill the numbers $1,2, \cdots, 36$ in a $6 \times 6$ grid, with each cell containing one number, such that the numbers in each row are in increasing order from left to right. Then the minimum sum of the six numbers in the third column is $\qquad$
4.63.

Let the six numbers filled in the third column, arranged in ascending order, be $A$, $B$, $C$, $D$, $E$, and $F$.

Since the row where $A$ is located needs to fill in two numbers smaller than $A$ before it, then $A \geqslant 3$; since the row where $B$ is located needs to fill in two numbers smaller than $B$, and both $A$ and the two numbers before $A$ are smaller than $B$, then $B \geqslant 6$.
Similarly, $C \geqslant 9$, $D \geqslant 12$, $E \geqslant 15$, $F \geqslant 18$.
Therefore, the sum of the six numbers filled in the third column is
$$
\begin{array}{l}
A+B+C+D+E+F \\
\geqslant 3+6+9+12+15+18=63 .
\end{array}
$$

Table 1 is one way to fill the numbers so that the sum of the six numbers in the third column reaches its minimum value (the filling method for the last three columns is not unique).