2. The ten positive integers from $1 \sim 10$ are written in a row in some order, denoted as $a_{1}, a_{2}, \cdots, a_{10}, S_{1}=a_{1}, S_{2}=a_{1}+a_{2}$, $\cdots, S_{10}=a_{1}+a_{2}+\cdots+a_{10}$. Then, among $S_{1}, S_{2}, \cdots, S_{10}$, the maximum number of primes that can occur is .
2. 7 .

Adding an odd number changes the sum to the opposite parity, and among even numbers, only 2 is a prime. Let $b_{i}$ be the $i$-th ($i=1,2,3,4,5$) odd number in this row. Then, when adding $b_{2}$ and $b_{4}$, the sums $S_{k}$ and $S_{n}$ are even numbers greater than 2. Therefore, $S_{k} \backslash S_{n}$ and $S_{10}=55$ must be composite numbers, meaning that in $S_{1}, S_{2}, \cdots, S_{10}$, the primes are no greater than 7. The example in Table 1 shows that there can be 7 primes.
Table 1
\begin{tabular}{|c|c|c|c|c|c|c|c|c|c|c|}
\hline$i$ & 1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 & 9 & 10 \\
\hline$a_{i}$ & 7 & 6 & 4 & 2 & 9 & 1 & 3 & 5 & 10 & 8 \\
\hline$S_{i}$ & 7 & 13 & 17 & 19 & 28 & 29 & 32 & 37 & 47 & 55 \\
\hline
\end{tabular}

Therefore, in $S_{1}, S_{2}, \cdots, S_{10}$, there can be at most 7 primes.