2. The inequality about $x$ $\left\{\begin{array}{l}x-1>0, \\ x^{3}-x^{2}+x \leqslant k\end{array}\right.$ has only the positive integer solutions 2 and 3. Then the value of $| k-21|+| k-52 \mid+$ 1975 is ( ).
(A) 2004
(B) 2005
(C) 2006
(D) 2007
2.C.

From the problem, we know that $x>1$.
Since $x=2, 3$ are solutions to $x^{3}-x^{2}+x \leqslant k$, we have,
$$
\left\{\begin{array}{l}
2^{3}-2^{2}+2 \leqslant k, \\
3^{3}-3^{2}+3 \leqslant k
\end{array}\right.
$$

Solving this, we get $k \geqslant 21$.
From the given information, $x=4,5,6, \cdots$ are solutions to $x^{3}-x^{2}+x>k$, so,
$$
\left\{\begin{array}{l}
4^{3}-4^{2}+4>k \\
5^{3}-5^{2}+5>k, \\
6^{3}-6^{2}+6>k, \\
\ldots \ldots
\end{array}\right.
$$

Solving this, we get $k<52$.
Therefore, $21 \leqslant k<52$.
Thus, the original expression is $(k-21)+(52-k)+1975=2006$.