Find the least positive integer $m$ such that $m^2 - m + 11$ is a product of at least four not necessarily distinct primes.
$m(m-1)$ is the product of two consecutive integers, so it is always even. Thus $m(m-1)+11$ is odd and never divisible by $2$. Thus any prime $p$ that divides $m^2-m+11$ must divide $4m^2-4m+44=(2m-1)^2+43$. We see that $(2m-1)^2\equiv -43\pmod{p}$. We can verify that $-43$ is not a perfect square mod $p$ for each of $p=3,5,7$. Therefore, all prime factors of $m^2-m+11$ are $\ge 11$.
Let $m^2 - m + 11 = pqrs$ for primes $11\le p \le q \le r \le s$. From here, we could go a few different ways:

Solution 1a
Suppose $p=11$; then $m^2-m+11=11qrs$. Reducing modulo 11, we get $m\equiv  1,0 \pmod{11}$ so $k(11k\pm 1)+1 = qrs$. 
Suppose $q=11$. Then we must have $11k^2\pm k + 1 = 11rs$, which leads to $k\equiv \mp 1 \pmod{11}$, i.e., $k\in \{1,10,12,21,23,\ldots\}$. 
$k=1$ leads to $rs=1$ (impossible)! Then $k=10$ leads to $rs=103$, a prime (impossible). Finally, for $k=12$ we get $rs=143=11\cdot 13$. 
Thus our answer is $m=11k= \boxed{132}$.

Solution 1b
Let $m^2 - m + 11 = pqrs$ for primes $p, q, r, s\ge11$. If $p, q, r, s = 11$, then $m^2-m+11=11^4$. We can multiply this by $4$ and complete the square to find $(2m-1)^2=4\cdot 11^4-43$. But 
\[(2\cdot 11^2-1)^2=4\cdot 11^4-4\cdot 11^2+1 <4\cdot 11^4-43<(2\cdot 11^2)^2,\]
hence we have pinned a perfect square $(2m-1)^2=4\cdot 11^4-43$ strictly between two consecutive perfect squares, a contradiction. Hence $pqrs \ge 11^3 \cdot 13$. Thus $m^2-m+11\ge 11^3\cdot 13$, or $(m-132)(m+131)\ge0$. From the inequality, we see that $m \ge 132$. $132^2 - 132 + 11 = 11^3 \cdot 13$, so $m = \boxed{132}$ and we are done.