Acording to what my friends told, the answer to a) its 0 so @StarLex1 its fakesolving once again...

StarLex1 wrote: 1....................12=(2020 digits of 1) and 1 digit of 2 and its reverse 211111.........11 it suppose to be an odd that end with 3 read carefully, all of the digit of the sum is odd

I think the method to solve this problem is to solve from the middle digit of the sum, and I'm doing by that way

For part a) We claim the answer is zero. Let $k_n$ be the $n$th digit of the $2021$-digit $k.$ It's clear that without regrouping, we want $(k_{1}+k_{2021}, k_{2}+k_{2020}, k_{3}+k_{2019}, \dots , k_{2021}+ k_{1}) \equiv 1 \pmod{2}.$ But note that $k_{1011}+k_{1011} \equiv 0 \pmod{2}.$ So we take into account regrouping to attempt to fix this. Notice that to cause regrouping after a sum of any arbitrary digits, we can just maximize both the digits while keeping their parity. Likewise, to "un-regroup" (i.e. do the reverse) we can just minimize both the digits while keeping their parity. Also note that since $k_{r}+k_{2022-r}$ appears twice in the integer (except for $r=1011$) we will have to account for this too by regrouping or changing parity in both instances. Now we rephrase the problem as follows, dividing into two cases. Case 1. $k$ plus its reverse forms a $2021$ digit number. That is, $k_1+k_{2021}$ does not regroup. The above has $2021$ digits $m_1, m_2, m_3, \dots m_{2021}.$ Notice that we can change the parity of both $m_r$ and $m_{2022-r},$ except when $r=1011$ (by changing the parity of $k_r$) or change the parity of both $m_{r}$ and $m_{2020-r}$ (by regrouping or "un-regrouping.") Since there can obviously exist $2021$ even digits (by $111...1+111...1,$ we start with that case. But now notice that when performing a move on odd indices of $m,$ we always change the parity of two odd indices of $m.$ Since there are $1011$ odd indices, it's impossible to change $1011$ even digits into $1011$ odd digits, so this case yields no guarani integers. Case 2. $k$ plus its reverse forms a $2021$ digit number. That is, $k_1+k_{2021}$ regroups. The above has $2022$ digits $1, m_1, m_2, m_3, \dots m_{2021}.$ Disregard the leading $1$ since it never changes, and only consider the $2021$ digit number $m_1, m_2, m_3, \dots m_{2021}.$ Notice that, similarly to Case 1, we can change the parity of both $m_r$ and $m_{2022-r}$ (except when $r=1011$) or change the parity of both $m_r$ and $m_{2020-r}$ (by regrouping or "un-regrouping.") We start with the simple case $999...9+999...9$ which yields $1999...98,$ or, disregarding the $1, 2020$ odd digits followed by the even digit $m_{2021}.$ But now, similarly to before, notice that when performing a move on odd indices of $m,$ we always change the parity of two odd indices of $m.$ Since there are $1011$ odd indices, it is impossible to change $1010$ odd digits and $1$ even digit into $1011$ odd digits, so this case yields no guarani integers as well, so we're done.

I'm the author of this problem Okay, so now the solution. Consider the number $N = \overline{a_{1} a_{2} \dots a_{n}}$, which is added to its reversed, $\overline{a_{n} a_{n-1} \dots a_{1}}$, whose result we call $b_{1} b_{2} \dots b_{n}$. Since we are interested only in the parity of the digits, we can take $\mod$ $2$ (if that makes easier the following definition). For every $i = 1, 2 \dots n$, define the pair $(c_i, d_i)$, where $c_i$ is $1$ iff $a_i - a_{n+1-i} \equiv 1 (mod$ $2)$, and $d_{i} = 1$ iff $a_{i} + a_{n+1-i} \leq 9$. We define $d_{0} = 1$. For all $b_{i}$'s to be $1$ ($mod$ $2)$, we need $c_{i} = d_{i-1}$, for all $i \geq 1$. By symmetry, we obtain $c_{i} = c_{n+1-i}$ and $d_{i} = d_{n+1-i}$, and with $c_{i} = d_{i-1}$, which implies $c_{n+1-i} = d_{n+2-i}$, and therefore $d_{i-1} = c_{i} = d_{i+1} = c_{i+2}$. As $n$ is odd, we write $n = 2k+1$, We have $c_{1} = 1$, in order for $b_{1}$ to be odd, and $c_{k+1} = 0$, as $a_{k+1} = a_{n+1 - (k+1)}$. If $k$ is even, then $k+1$ is odd, so $1 = c_{1} = c_{k+1} = 0$. Contradiction If $k$ is odd, then, for $i$ odd, we have $(c_{i}, d_{i}) = (1, 0)$, and for $i$ even, $(c_{i}, d_{i}) = (d_{i-1}, c_{i+1}) = (0, 1)$. There are $20$ ordeder pairs of digits $(a_{i}, a_{n+1-i})$ which satisfy the first condition ($i$ odd), and $25$ pairs which satisfy the second condition. So, for $a)$, we note that its corresponding $k$ is even, so there are no guarani numbers. For $b)$, by the multiplicative principle, and $2023 = n = 2 \times 1011 + 1$, we obtain $20$ choices which generate $(c_1, d_1)= (1, 0)$, 25 choices which generate $(c_2, d_2) = (0, 1)$ $\dots$ $20$ choices which generate $(c_k, d_k) = (1, 0)$, and finally, $5$ choices for $(c_{k+1}, d_{k+1}) = (0, 1)$. The final answer must be therefore the product of all these posibilities, that is, $$ 100 \times 500^{505} $$