, 2026

Suppose the number of complete strings of $1234$ is $k$. We can compute the number of digits up
to and including the rightmost $4$ in
the string. It is $4k$ ($k$ occurrences of each of $1$, $2$, $3$, and $4$) plus the number of $5$s, which is $$4k + \bigg(1+2+3+4+\cdots+(k-2)+(k-1)\bigg)$$ Note that the $k$ 5s that follow the final $1234$ are not included in this count. This
sum is equal to $$4k+\dfrac{(k-1)k}{2}=\dfrac{8k+k^2-k}{2} = \dfrac{k^2+7k}{2}$$ so there are $\dfrac{k^2+7k}{2}$ digits up to and
including the rightmost $4$.

The total number of digits is $2026$,
so $k$ must be the greatest integer
satisfying $\$\dfrac{k^2+7k}{2} \leq$
2026$, which is equivalent to$$k^2+7k\leq$ 4052$.

Notice that $60^2+7(60)=4020$ but
$61^2+7(61)=4148$, so we have that
$k=60$, meaning that there are $60$ complete strings of $1234$.

Thus, the number of digits up to and including the rightmost $4$ is $\$\dfrac{60^2+7(60)}{2} = \dfrac{4020}{2}=2010$$. There would be$60$5s after the$60$th$4$, which is more than the$2026-2010=16$digits needed to reach$2026$ digits in total. This means there
are no "partial" strings of $1234$.

Therefore, among the first $2026$
digits, there are $60$ each of $1$, $2$, $3$, $4$
for a total of $4\times 60=240$
digits, and the rest must be $5$.
Thus, $2026-240=1786$ of the digits
are $5$.

The sum of the first $2026$ digits
is $$S=1\times 60 + 2\times 60 + 3\times 60 + 4\times 60 + 5\times 1786 = 9530$$ The sum of the digits of $S$ is $9+5+3+0=17$.

Suppose the number of complete strings of $1234$ is $c$. We can compute the number of digits up
to and including the rightmost $4$ in
the string. It is $4c$ ($c$ occurrences of each of $1$, $2$, $3$, and $4$) plus the number of $5$s, which is $$4c + \bigg(1+2+3+4+\cdots+(c-2)+(c-1)\bigg)$$ Note that the $c$ 5s that follow the final $1234$ are not included in this count. This
sum is equal to $$4c+\dfrac{(c-1)c}{2}=\dfrac{8c+c^2-c}{2} = \dfrac{c^2+7c}{2}$$ so there are $\dfrac{c^2+7c}{2}$ digits up to and
including the rightmost $4$.

The total number of digits is $2026$,
so $c$ must be the greatest integer
satisfying $\$\dfrac{c^2+7c}{2} \leq$
2026$, which is equivalent to$$c^2+7c\leq$ 4052$.

Notice that $60^2+7(60)=4020$ but
$61^2+7(61)=4148$, so we have that
$c=60$.

Thus, the number of digits up to and including the rightmost $4$ is $\$\dfrac{60^2+7(60)}{2} = \dfrac{4020}{2}=2010$$. There would be$60$5s after the$60$th$4$, which is more than the$2026-2010=16$digits needed to reach$2026$ digits in total. This means there
are no "partial" strings of $1234$.

Therefore, among the first $2026$
digits, there are $60$ each of $1$, $2$, $3$, $4$
for a total of $4\times 60=240$
digits, and the rest must be $5$.
Thus, $2026-240=1786$ of the digits
are $5$.

The sum of the first $2026$ digits
is $$S=1\times 60 + 2\times 60 + 3\times 60 + 4\times 60 + 5\times 1786 = 9530$$ The sum of the digits of $S$ is $9+5+3+0=17$.