Annual Harvard-MIT Mathematics Tournament

Solution:
The main idea is the following lemma:

Lemma. For any non-negative integer $n$ and prime $p$, $$\sum_{k=n+1}^{n+p^{2}-p} k^{k} \equiv 1 \pmod{p}.$$

Proof. Note that $a^{b}$ depends only on the value of $a \pmod{p}$ and the value of $b \pmod{p-1}$. Since $p$ and $p-1$ are relatively prime, the Chinese Remainder Theorem implies that any $p^{2}-p$ consecutive integers will take on each possible pair of a residue $\bmod\ p$ and a residue $\bmod\ p-1$. In other words, if we let $(a, b) = (k \bmod p, k \bmod p-1)$, then as $k$ ranges through $p^{2}-p$ consecutive integers, $(a, b)$ will range through all $p^{2}-p$ ordered pairs of residues $\bmod\ p$ and residues $\bmod\ p-1$. This implies that
$$\sum_{k=n+1}^{n+p^{2}-p} k^{k} \equiv \sum_{b=1}^{p-1} \sum_{a=1}^{p} a^{b} .$$
It is well-known that $\sum_{a=1}^{p} a^{b} = \begin{cases} -1 & p-1 \mid b \\ 0 & p-1 \nmid b \end{cases}$. We will sketch a proof here. When $p-1 \mid b$, the result follows from Fermat's Little Theorem. When $p-1 \nmid b$, it suffices to consider the case when $b \mid p-1$, since the $b$th powers $\bmod\ p$ are the same as the $\gcd(b, p-1)$th powers $\bmod\ p$, and there are an equal number of every non-zero $b$th power. But in this case, the $b$th powers are just the solutions to $x^{\frac{p-1}{b}} - 1$, which add up to zero by Vieta's formulas.

Now, using the formula for $\sum a^{b}$, we get that
$$\sum_{b=1}^{p-1} \sum_{a=1}^{p} a^{b} \equiv -1 \pmod{p}$$
which completes the lemma.

We now apply the lemma with $p=101$ and $n=3, 10103$, and $20103$ to get that $\sum_{k=1}^{30303} k^{k} \equiv \left(\sum_{k=1}^{3} k^{k}\right) - 3$. But $\sum_{k=1}^{3} k^{k} = 1^{1} + 2^{2} + 3^{3} = 1 + 4 + 27 = 32$, so the answer is $32 - 3 = 29$.