Ummmm.... maybe I am missing something but if we make $ n $ sufficiently large, then $ F_n $ will have more than $ c $ digits in base $ b $ and so will satisfy the desired condition. Am I interpreting the problem correctly?

Ummmm... maybe I'm missing something, but if we make $n$ sufficiently large, then $10^n$ will have more than $c$ digits in base $10$ and so will satisfy the desired condition.

The problem says "the sum of the digits of $F_n$". The word to note is "sum".

The Fibonacci sequence contains multiples of any positive integer. Take a multiple of $U = \underbrace{11\dots 1}_{c+1\text{ times}}$ in base $b$. It is well known such a multiple has digit sum at least $c + 1$ is base $b^{(*)}$. $(*)$ Proof: Let such multiple be $M$. While $M \ge b^{c + 1}$ choose any digit $d \neq 0$ at position $k \ge c + 1$ (0-based, from the right), replace $M$ by $M' = M - db^k + db^{k - (c + 1)}$. This operation preserves divisibility by $U$, and digit sum either stays the same or decreases due to carry. In the end we obtain an $M' < b^{c + 1}$ which is multiple of $U$ because $M$ initially was, hence $M' = \underbrace{dd\dots d}_{c+1\text{ times}}$ for some $1\le d \le b - 1$, so the digit sum of $M'$ is at least $c + 1$, and therefore the digit sum of $M$ initially also was at least $c+1$.

Sorry if I sound silly, but can't we just select a Fibonacci number which is a multiple of $b^{\text{something large}}-1$?

The problem is a cute one, although far too easy for the ELMO shortlist. Since the sequence cycles modulo anything, pick some massively large $k$, larger than $c$ should work. Now the sequence achieves $F_i\equiv F_{i+1}\equiv 1\pmod{b^k}$ infinitely many times. Pick any other than the first, and work backwards. $F_{i-1}\equiv 0, F_{i-2}\equiv 1, F_{i-3}\equiv b^k-1\pmod{b^k}$. So $F_{i-3}$ has $k$ copies of the digit $b-1$ (plus whatever at the beginning) and obviously $k(b-1)\ge k>c$ for large $k$.

The problem needs just a sentence : $a_{-2}=-1$

Pretty easy. It is well known that $F_i$ is strictly periodic mod $n$, to say that there exists large $N$ where $F_N \equiv 0$ and $F_{N+1} \equiv 1$ (mod $n$). Thus, $F_{N-1} \equiv 1$ (mod $n$), implying that $F_{N-2} \equiv -1$ (mod $n$). Taking $n=b^k$ for arbitrarily large $k$ implies that we can find some $\ell$ where $F_\ell \equiv -1$ (mod $b^k$). Thus, $F_\ell$ will have $k$ digits of value $b-1$ at the end. Implying that \[s_b(F_\ell) \geq k(b-1) \] But $k(b-1)>c$ for sufficiently large $k$.

v_Enhance wrote: Define the Fibanocci sequence recursively by $F_1=1$, $F_2=1$ and $F_{i+2} = F_i + F_{i+1}$ for all $i$. Prove that for all integers $b,c>1$, there exists an integer $n$ such that the sum of the digits of $F_n$ when written in base $b$ is greater than $c$. Proposed by Ryan Alweiss Posting for storage:- It is pretty well know that $f_n$ is periodic modulo $i$ When $k \rightarrow \inf$,then by Kronecker's Theorem pick an integer $\mathbf{L}$ such that $f_L \equiv 1 \pmod {b^k}$ and the last $k$ digits of $F_l$ are all $b-1$ so \[s_b(F_L) \geq k(b-1) \],and we are done.

Maybe it is possible to only use density.