Answer: $\quad 229$
Solution: Say $n$ is $p$-good if $\sum_{k=1}^{n}(-1)^{\nu_{p}(k!)}p^{2}$, so we must have $n+1>p^{3}$.
Checking $p=3$, the smallest potential $n+1$ 's are
- $2 \cdot 3^{3}-2 \cdot 3=48$, which does not have a prime factor $q>3$.
- $2 \cdot 3^{3}=54$, which does not have a prime factor $q>3$.
- $2 \cdot 3^{3}+2 \cdot 3=60$, which does not work because 60 is the wrong size for $q=5$.

The next value $2 \cdot 3^{5}-2 \cdot 3^{3}-2 \cdot 3$ is already bigger than 230 .
Checking $p=5$, the smallest potential $n+1$ 's are
- $2 \cdot 5^{3}-4 \cdot 5=230$, which works for $q=23$.

For $p \geq 7, n+1 \geq p^{3}>230$, so 229 is the smallest value of $n$.
It suffices to prove the claim. We argue via a series of lemmas. We introduce the notation of $S(a, b)=$ $\sum_{k=a}^{b-1}(-1)^{\nu_{p}(k!)}$. Note that $n$ is $p$-good if and only if $S(0, n+1) \leq 0$.
Lemma 1. If $n \leq p^{2}, S(0, n)$ is the distance to the nearest even multiple of $p$.
Proof. This follows straightforwardly from the fact that $\nu_{p}(k!)=\lfloor k / p\rfloor$ for $n \leq p^{2}$.
Lemma 2. If $a$ and $b$ are positive integers so that $b \leq p^{\nu_{p}(a)}$, then $S(a, a+b)=(-1)^{\nu_{p}(a!)} S(0, b)$.
Proof. Note that for $00$ and suppose $n=a p^{2}+b$ for $0 \leq b<p^{2}$. Lemma 6 tells us that $S(0, n)=$ $p S(0, a)+(-1)^{\nu_{p}(a)} S(0, b)$. Since $0 \leq S(0, b) \leq p$ (by Lemma 1), the only way for $S(0, n)$ to be less than 0 is if $S(0, a)=0$ and $(-1)^{\nu_{p}(a)}=-1$, which is impossible since $P(a)$ holds.
There are two ways for $S(0, n)=0$ to be true. The first case is that $S(0, a)=S(0, b)=0$, which implies by $P(a)$ and Lemma 1 that $a$ is neat and $b$ is a multiple of $2 p$. This captures the neat numbers with $a_{0} \geq 0$. Note that in this case $\nu_{p}(n!)=\nu_{p}\left(\left(a p^{2}\right)!\right)+\nu_{p}(b!)$ (by the same logic as Lemma 2), which is even as $\nu_{p}\left(\left(a p^{2}\right)!\right)$ is even by $P(a)$ and Lemma 3 and $\nu_{p}(b!)=b / p$ which is even.
The second way for $S(0, n)$ to be 0 is if $S(0, a)=1, S(0, b)=p$, and $(-1)^{\nu_{p}(a!)}=-1$. By the inductive hypothesis, this is equivalent to $a+1$ being neat and $b$ being an odd multiple of $p$. This captures exactly the neat numbers with $a_{0}<0$. Also, $\nu_{p}(n!)=\nu_{p}\left(\left(a p^{2}\right)!\right)+\nu_{p}(b!)$, which is even as $\nu_{p}(a!)$ is odd and $\nu_{p}(b!)$ is odd.
Analyzing the possibilities where $S(0, n)=1$ is almost exactly the same as the above, so we will omit it here. We encourage the reader to fill in the details.
10. Compute the smallest positive integer $n$ for which there are at least two odd primes $p$ such that
$$
\sum_{k=1}^{n}(-1)^{\nu_{p}(k!)}<0 .
$$

Note: for a prime $p$ and a positive integer $m, \nu_{p}(m)$ is the exponent of the largest power of $p$ that divides $m$; for example, $\nu_{3}(18)=2$.