Step 1: Notation.
Let the cells in the empty (fourth) column from top to bottom be
\[
a=\text{(row1,col4)},\quad b=\text{(row2,col4)},\quad c=\text{(row3,col4)},\quad d=\text{(row4,col4)},
\]
where each takes value $1$ if it is a cross (X), else $0$.
Step 2: Use numbered cells that touch the empty column.
- Cell $(1,3)$ has number $2$ and neighbors $\{(1,4)=a,\ (2,4)=b,\ (2,2)=X\}$ among crosses around it. Since $(2,2)$ is already a cross,
\[
a+b+1=2 \;\Rightarrow\; a+b=1. \tag{1}
\]
- Cell $(2,3)$ has number $3$ and neighbors including two fixed crosses $(2,2)$ and $(3,2)$, and the fourth-column cells $a,b,c$. Hence
\[
a+b+c+2=3 \;\Rightarrow\; a+b+c=1 \;\Rightarrow\; c=0 \quad\text{(using (1)).} \tag{2}
\]
- Cell $(3,3)$ has number $4$ and neighbors include three fixed crosses $(2,2)$, $(3,2)$, $(4,3)$ and the fourth-column cells $b,c,d$. Thus
\[
b+c+d+3=4 \;\Rightarrow\; b+d=1. \tag{3}
\]
Step 3: Maximize crosses in the fourth column.
We want to maximize $a+b+c+d$. From (2), $c=0$. From (1) and (3):
\[
a=1-b,\qquad d=1-b.
\]
Therefore
\[
a+b+c+d=(1-b)+b+0+(1-b)=2-b,
\]
which is maximized when $b=0$. Then $a=1,\ c=0,\ d=1$, yielding
\[
a+b+c+d=2.
\]
\[
\boxed{2}
\]