We use the following observation: Any odd numbers $x_1, \dots, x_k$ satisfy
$$x_1 x_2 + x_2 x_3 + \dots + x_{k-1} x_k + x_k x_1 \equiv k \pmod{4}. \quad (*)$$
Note that the sum in $(*)$ is cyclic, unlike the ones in the statement. To justify $(*)$ reduce the $x_i$ mod $4$; then they become $+1$'s or $-1$'s as odd numbers are congruent to $\pm 1$ mod $4$. Note that replacing an $x_i = -1$ by $x_i = 1$ does not change the mod $4$ remainder of $S = \sum_{j=1}^{k} x_j x_{j+1}$. Indeed the new and the old value of $S$ differ by $2(x_{i-1} + x_{i+1})$ which is a multiple of $4$ as $x_{i-1}, x_{i+1}$ are odd. So we may assume $x_i = 1$ for all $i$, then $S = k$ and $(*)$ is obvious.

Let $d_1 d_2 \dots d_{2013}$ satisfy the stated conditions. By $(*)$ we have $\sum_{j=1}^{1810} d_j d_{j+1} \equiv 1810 \pmod{4}$ (here

d_{1810} + 1 = d_1), hence $\sum_{j=1}^{1809} d_j d_{j+1} \equiv 1 \pmod{4}$ if and only if $1810 - d_{1810} d_1 \equiv 1 \pmod{4}$, i.e.
$d_1 d_{1810} \equiv 1 \pmod{4}$.

Similarly $\sum_{j=1810}^{2012} d_j d_{j+1} \equiv 1 \pmod{4}$ if and only if $(2013-1809) - d_{1810} d_{2013} \equiv 1 \pmod{4}$, i.e.
$d_{1810} d_{2013} \equiv -1 \pmod{4}$. We see that the conditions depend only on the three digits $d_1, d_{1810}, d_{2013}$; the
remaining $2010$ digits $d_i$ can be chosen arbitrarily among $1, 3, 5, 7, 9$.

There are $3$ odd decimal digits $\equiv 1 \pmod{4}$, namely $1, 5, 9$; there are $2$ odd digits $\equiv -1 \pmod{4}$,
namely $3, 7$. Let $d_{1810} \in \{1,5,9\}$. Then $d_1 d_{1810} \equiv 1 \pmod{4}$ and $d_{1810} d_{2013} \equiv -1 \pmod{4}$ imply
$d_1 \in \{1,5,9\}, d_{2013} \in \{3,7\}$. So there are $3$ choices for each of $d_1$ and $d_{1810}$, and $2$ choices for $d_{2013}$.
The choices are independent, which gives $3 \cdot 3 \cdot 2 = 18$ admissible choices for the triple $d_1, d_{1810}, d_{2013}$.
Because there are $5$ choices for each of the remaining $2010$ digits $d_i$ (they can be arbitrary), we
obtain $18 \cdot 5^{2010}$ admissible numbers $d_1 d_2 \dots d_{2013}$ with $d_{1810} \in \{1,5,9\}$.

Likewise if $d_{1810} \in \{3,7\}$ then $d_1 \in \{3,7\}$, $d_{2013} \in \{1,5,9\}$. Thus there are $2$ choices for each of $d_1$ and
$d_{1810}$, and $3$ choices for $d_{2013}$, leading to $2 \cdot 2 \cdot 3 = 12$ admissible choices for the triple $d_1, d_{1810}, d_{2013}$.
Like in the previous case we obtain $12 \cdot 5^{2010}$ admissible numbers with $d_{1810} \in \{3,7\}$.

In summary there are $18 \cdot 5^{2010} + 12 \cdot 5^{2010} = 6 \cdot 5^{2011}$ admissible numbers in all.