11. The smallest positive integer $n$ for which $n^{2}-n+11$ has four prime factors (not necessarily distinct) is $\qquad$

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11.132.

From Table 1, it can be verified that for any positive integer $n, n^{2}-$ $n+11$ is not a multiple of $2,3,5,7$.
Table 1
\begin{tabular}{|c|c|c|c|c|c|c|c|c|c|c|c|c|c|c|c|c|c|}
\hline & $\bmod 2$ & $\bmod 3$ & \multicolumn{3}{|c|}{$\bmod 5$} & \multicolumn{3}{|c|}{$\bmod 7$} \\
\hline$n$ & 0 & 1 & 0 & 1 & 2 & 0 & 1 & 2 & 3 & 4 & 0 & 1 & 2 & 3 & 4 & 5 & 6 \\
\hline$n^{2}$ & 0 & 1 & 0 & 1 & 1 & 0 & 1 & 4 & 4 & 1 & 0 & 1 & 4 & 2 & 2 & 4 & 1 \\
\hline$n^{2}-n+11$ & 1 & 1 & 2 & 2 & 1 & 1 & 1 & 3 & 2 & 3 & 4 & 4 & 6 & 3 & 2 & 3 & 6 \\
\hline
\end{tabular}

Therefore, the smallest possible value is $n^{2}-n+11=11^{4}$.
Then $n(n-1)=2 \times 5 \times 7 \times 11 \times 19$.
However, this equation has no positive integer solutions.
When $n^{2}-n+11=13 \times 11^{3}$, we get
$n(n-1)=2 \times 2 \times 3 \times 11 \times 131$.
Solving this, we find $n=132$.