International Mathematical Olympiad

First, we prove that this can always be achieved. Without loss of generality, suppose at least $\frac{2022}{2}=1011$ terms of the $\pm 1$-sequence are $+1$. Define a subsequence as follows: starting at $t=0$, if $a_t=+1$ we always include $a_t$ in the subsequence. Otherwise, we skip $a_t$ if we can (i.e. if we included $a_{t-1}$ in the subsequence), otherwise we include it out of necessity, and go to the next $t$. Clearly, this subsequence will include all $+1$s. Also, for each $-1$ included in the sequence, a $-1$ must have been skipped, so at most $\lfloor \frac{1011}{2}\rfloor =505$ can be included. Hence the sum is at least $1011-505=506$, as desired.

Next, we prove that, for the $\pm 1$-sequence $$(\{-1\},\{+1,+1\},\{-1,-1\},\{+1,+1\},\dots ,\{+1,+1\},\{-1,-1\},\{+1\})$$ each admissible subsequence $a_{t_i}$ has $-506⩽{\sum }_i{a}_{t_i}⩽506$. We say that the terms inside each curly bracket is a block. In total, there are $1012$ blocks - $506$ of them hold $+1$s, and $506$ of them hold $-1$s. (The two blocks at each end hold $1$ number each, each other block holds $2$.)

Suppose an admissible subsequence includes terms from $k$ blocks holding $+1$s. Then, in each $-1$-pair in between the $+1$-pairs, the subsequence must also include at least one $-1$. There can be at most two $+1$s included from each $+1$-block, and at least one $-1$ must be included from each $-1$-block, so the sum is at most $2k-(k-1)=k+1$.

For $k<506$, this is at most $506$. If $k=506$, one of the $+1$-blocks must be the one at the end, meaning it can only include one $+1$, so that the maximum in this case is only $k$, not $k+1$, so in this case the sum is also at most $506$.

Hence we have shown that for any admissible subsequence, ${\sum }_i{a}_{t_i}⩽506$. Analogously we can show that $-506⩽{\sum }_i{a}_{t_i}$, meaning that $C⩽506$ as desired.