Solution:

Since $\cos(\pi k) = (-1)^k$ and $k^2 + 3k + 2 = (k+1)(k+2),$ we have $f(k) = \left[(k+1)(k+2)\right]^{(-1)^k}.$ The sum of the logarithms is the log of the product $\prod_{k=1}^n f(k),$ which telescopes: consecutive factors $\frac{1}{(k+1)(k+2)}$ and $(k+2)(k+3)$ leave only boundary terms.

For even $n$ the product is $\frac{3 \cdot 4 \cdots (n+2)}{2 \cdot 3 \cdots (n+1)} = \frac{n+2}{2},$ and for odd $n$ it is $\frac{3 \cdot 4 \cdots (n+1)}{2 \cdot 3 \cdots (n+2)} = \frac{1}{2(n+2)}.$ The absolute value of the log equals $1$ exactly when the product is $10$ or $\frac{1}{10}.$

For even $n,$ $\frac{n+2}{2} = 10$ gives $n = 18;$ for odd $n,$ $2(n+2) = 10$ gives $n = 3.$ The requested sum is $18 + 3 = 21.$