The answer is $233625$.

Suppose that the centers of unit squares have coordinates $(i,j)$, where $i = 1,2,\ldots,2013$; $j = 1,2,\ldots,2013$. The unit square with center at $(i,j)$ will be denoted by $u(i,j)$. Let the marked unit squares be:
$u(19k, 19l + 1)$, $u(19k, 19l + 2)$, where $1 \le k \le 105$, $0 \le l \le 105$ and
$u(m, 19n)$, where $1 \le m \le 2013$, $1 \le n \le 105$.
Then it can be readily seen that the total number of marked unit squares is $233625$, and any sub-square $19 \times 19$ has exactly $21$ marked unit squares.
Let $k \ge 2$ be a positive integer. Now by the method of mathematical induction we show that if any $19 \times 19$ sub-square of the grid $(19k-1) \times (19k-1)$ has at least $21$ marked unit squares, then the total number of marked unit squares is at least $M(k) = (k-1)(21k-1)$.

* $k = 2$. $M(2) = 41$. Consider two $19 \times 19$ squares: the square consisting all $u(k,l)$, where $1 \le k \le 19$, $1 \le l \le 19$ and the square consisting all $u(k,l)$, where $19 \le k \le 37$, $19 \le l \le 37$. Each of these $19 \times 19$ squares contains at least $21$ marked unit squares and their intersection is the unit square $u(19, 19)$. Therefore the total number of marked unit squares is at least $21 + 21 - 1 = 41$. Done.

* Suppose the statement is correct for a $((19k-1) \times (19k-1))$ grid $A$ and consider a $((19(k+1)-1) \times (19(k+1)-1))$ grid $B$. Suppose that $A$ consists of all unit squares $u(i,j)$, where $1 \le i \le 19k+18$, $1 \le j \le 19k+18$ and $B$ consists of all unit squares $u(i,j)$, where $1 \le i \le 19k+18$, $1 \le j \le 19k+18$.
Let $19 \times 19$ squares $U_s$, $s = 1,2,\ldots,k$; consist of all unit squares $u(i,j)$, where $19k \le i \le 19k+18$, $19s-18 \le j \le 19s$ and $19 \times 19$ squares $V_t$, $t = 1,2,\ldots,k$; consist of all unit squares $u(i,j)$, where $19t-18 \le i \le 19t$, $19k \le j \le 19k+18$. Note that the squares $U_k$ and $V_k$ share a unit square $u(19k-1, 19k-1)$, all other pairs of $U_s$ and $V_t$ squares do not share any unit square. Therefore, since the union of $k$ $U_s$ and $k$ $V_t$ squares is a subset of the set $B-A$, the set $B-A$ contains at least $21 \cdot 2k-1 = 42k-1$ marked squares. Thus, by inductive hypothesis $B$ contains at least $=(k-1)(21k-1)+42k-1 = ((k+1)-1)(21(k+1)-1)$. Done. Since $M(106) = 233625$ the solution is completed.