This amazing problem comes from the Balkan MO 2022 Shortlist. Specifically, it is C5. This solution follows along the first official solution given for this wonderful problem. Call a position of the added cubelet $k-$good if the resulting solid can be covered with $k \times 1 \times 1$ cuboids . Partition the $2021 \times 2021 \times 2021$ cube into $2021^3$ unit cubes, and to each unit cube with coordinates $(x,y,z)$ assign the number $\omega^{x+y+z}$, where $\omega$ is a primitive $k-$th root of unity, and call this number the value of the unit cube. Let $(a,b,c)$ be the coordinates of the added unit cube. Note that if $(a,b,c)$ is $k-$good, then $(2020-a,b,c)$ is $k-$good too, due to symmetry. Now, let the value of a solid be the sum of the the values of the cubes it consists of. We have the following Lemma: Lemma: The value of any $k \times 1 \times 1$ cuboid is $0$. Proof: Since $\omega$ is a $k-$th root of unity, if for example the coordinates of the cubes the cuboid consists of are $(m+i,n,p)$ for $i \in \{0,1,\ldots,k-1\}$, then $\sum_{i=0}^{k-1} \omega^{m+i+n+p}=\omega^{m+n+p} \sum_{i=0}^{k-1} \omega^i=0,$ as desired $\blacksquare$ Claim: $k \mid 2022$. Proof: The value of the whole $2021 \times 2021 \times 2021$ cube is obviously $\displaystyle (\sum_{i=0}^{2020} \omega^i)^3,$ and so by our Lemma we must have that $\displaystyle (\sum_{i=0}^{2020} \omega^i)^3=-\omega^{a+b+c}$. Note that $\displaystyle \sum_{i=0}^{2020} \omega^i=\dfrac{1-\omega^{2021}}{1-\omega},$ and since $|\omega^{a+b+c}|=|\omega|^{a+b+c}=1,$ we must have $|1-\omega^{2021}|=|1-\omega|$. Let $\omega=\cos \dfrac{2\pi}{k}+i\sin \dfrac{2\pi}{k}$ (note that $\omega$ is a primitive root of unity). Then, using De Moivre's formula, $|1-\omega^{2021}|=|(1-\cos \dfrac{2021 \cdot 2\pi}{k})+i\sin \dfrac{2021 \cdot 2\pi}{k}|$ and $|1-\omega|=|(1-\cos \dfrac{2\pi}{k})+i\sin \dfrac{2\pi}{k}|,$ and so we obtain that $(1-\cos \dfrac{2021 \cdot 2\pi}{k})^2+(\sin \dfrac{2021 \cdot 2\pi}{k})^2=(1-\cos \dfrac{2\pi}{k})^2+(\sin \dfrac{2\pi}{k})^2,$ which easily rewrites as $\cos \dfrac{2021 \cdot 2\pi}{k}=\cos \dfrac{2\pi}{k}$. Therefore, $\dfrac{2021 \cdot 2\pi}{k}=2\ell\pi \pm \dfrac{2\pi}{k}$ for some positive integer $\ell$. Therefore, $2020=\ell k$ or $2022=\ell k$, and so $k \mid 2020$ or $k \mid 2022$. Moreover, note that $k$ must divide $2021^3+1$, the total number of unit cubes in the new solid. Therefore, if $k \mid 2020$, then $k \mid 1^3+1=2,$ and so $k \mid 2$. However, if $k \mid 2$ then trivially $k \mid 2022$, and so in any case we may suppose that $k \mid 2022$, as desired $\blacksquare$ Now, since $k \mid 2022$, we infer that $\displaystyle \sum_{i=0}^{2021} \omega^i=0,$ and so we have $\omega^{a+b+c}=-\displaystyle (\sum_{i=0}^{2020} \omega^i)^3=\omega^{3 \cdot 2021},$ therefore $k \mid (a+b+c)-3 \cdot 2021,$ and since $2021 \equiv -1 \pmod k$, we obtain that $a+b+c \equiv -3 \pmod k$. Hence, we infer that if $(a,b,c)$ is $k-$good, then $a+b+c \equiv -3 \pmod k$. Due to our previous observation, since $(2020-a,b,c)$ is $k-$good too, we obtain that $2020-a+b+c \equiv -3 \pmod k,$ and so $b+c-a \equiv -1 \pmod k$, hence $2a \equiv -2 \pmod k$. WLOG we may assume that the added unit cube is at the bottom of the initial cube, and so $c=-1$. Therefore, $2a \equiv -2 \pmod k$ and $b \equiv a \pmod k$. Note that if a triple $(a,b,c)$ is $k-$good, then it is also $p-$good for any prime $p \mid k$, and so it suffices to only consider $k \in \{2,3,337 \}$. Note that if $k=2$, the conditions above imply that $a \equiv b \pmod 2$, and if $k \in \{3,337 \}$, the conditions imply that $a \equiv b \equiv -1 \pmod k$. We will in fact prove that these necessary conditions are sufficient, too. Proof of sufficiency: Firstly let $k=2$. It is easy to see that if we remove the bottom $2021 \cdot 2021 \cdot 1$ square of unit cubes plus the added cubelet, the remainder is a $2021 \cdot 2021 \cdot 2020$ solid, which can trivially be covered with $2 \times 1 \times 1$ cubeletes. To cover the removed solid, note that we may place a $2 \times 1 \times 1$ cubelet covering the added unit cube and the cube exactly above it, and now we are equivalently left to cover a $2021 \cdot 2021$ square with a unit square removed with $2 \times 1$ dominoes, which is easy by partitioning this shape into rectangles whose borders are this square. It is easy to see that this construction works, due to the condition $b \equiv a \pmod 2$. Now, if $k \in \{3,337 \}$, we may apply the above construction again, this time firstly covering a $2021 \times 2021 \times (2022-k)$ solid, then using a $k \times 1 \times 1$ cubelet to cover the added unit cube and the $k-1$ unit cubes above it, and lastly being left with $k-1$ copies of a $2021 \times 2021$ square with one square removed, which is again covered as described before. Having proved the sufficiency of the conditions we obtained, we are just left with the tedious computation part. We want to count the number of pairs $(a,b)$ such that $0 \leq a,b \leq 2020$ and either $a \equiv b \pmod 2$, or $a \equiv b \equiv -1 \pmod 3$, or $a \equiv b \equiv -1 \pmod {337}$. I will omit this part since it is quite boring and standard though For reference, the shortlist packet gives a final answer of $13612182$.

For anyone wondering, this question is actually not that hard. Firstly, we can discard all composites because if they work for some position, so does one of its primes. The roots of unity argument is just colouring in disguise, when we colour the cube mod $k$ such that any 1x1x$k$ takes each colour once, we need all the colours to appear an equal number of times except for one that appears one less time. Doing this counting is classic with roots of unity separating the mod $k$ values. This will show $k\mid 2022$. We just consider $k=2,3,337$. To find the actual possible locations, consider the same colouring but reflected across each direction. This gives even more constraints which turns out to be sufficient. The answer extraction though... horrendous.

I don’t know what you guys think you’re doing, especially #1 and #2 (who seem to know the source of the problem) since the BMO 2022 Shortlist must be kept confidential until BMO 2023. Somebody please report this thread so a mod can close it (I cannot do it from my phone).

Let me write a post in defence of #2. The solution of this problem is publicly visible (since 1-st of April) on a certain serbian math website (where the MO problems can be found), so I don't think posting roughly the same solution as the one given in the official Serbian MO solutions is a problem, especially having in mind that the BMO is to be held next week and I really doubt that there is a country that selects their team a week before the contest (and according to the official BMO website the deadline to register the participants is 26-th of April, which has passed). For sure, discussing the BMO SL before the BMO is not the best idea, but in this case it's not that much of a problem.

@oVlad Vlad, the problem is available (in Serbian) here, since at least April 3rd (when I PMed a_507_bc asking whether he could post problems 2 and 5 from 2023 Serbia MO, too, and he gave me the aforementioned link). I learned that this problem comes from the BMO Shortlist quite later, about a week or two ago. Let's not be hypocrites, please. Dozens of people have visited that publicly available site, and even more have seen it anyway since the shortlist is not a CIA document. If post #1 was made a month ago, I would've probably reported it or PMed the OP (although, technically, he did nothing illegal. The bad handling comes from the Serbia Math Society for allowing the problem to be published). But now, it is 1st of May. All TSTs are over. Registration for the BMO has closed. I really don't see why this thread is bound for report. I can even be as brave (for some definition of bravery) as to say that if my posting of a solution disrupts any TST procedure, I will be more than happy to walk away from being a member of the Greak team at this year's BMO. Jesus Christ, that essay is college applications material, I should include it next year at my applications.

Dear #6 and #7 posters, You’re right, nothing bad will happen, there are no more TSTs and this won’t affect anything. However, even if the problem was already shared on some website, this doesn’t mean that you should go around re-posting it. It’s more or less the same logic as “if this guy throws an empty bottle on the lawn in the park, surely I can do that too, right?”. It’s just a matter principle. I really don’t mean to be a hypocrite (and you know how much I care about this specific topic Orestis), I just want to follow the rules, especially when it comes to such an important event. But then again, the BMO or their rules might not be important to some of you. Edit: I am sorry if I was rude before, I’m glad we talked things out. See you in Turkey

@oVlad Firstly, let me apologize for implicitly calling you a hypocrite. I was quite harsh, I'm sorry. Secondly, I understand that you want to go strictly by the rules, and so do I, but we just have a slightly different way of thinking. It goes without saying that the rules and the BMO itself are very important for all of us, and we always try to show uttermost respect to them. By the way: incidentally, I have once posted a BMO Shortlist problem before the discussion embargo was over, but to my defence I had no idea back then it was from the shortlist See you in Turkey next week!