You should really make this question more clear. What do you mean by a "partial resolution"? For instance, is it normal, or at least $S_2$? 
Is it $\mathbb Q$-factorial? Or is at least the canonical divisor $\mathbb Q$-Cartier? You need some assumption for the question to make sense.

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First let's take $n=2$. Blowing up the singular point gives you the crepant resolution $\pi:Y\to X$. (By the way if a surface has a crepant resolution it must be the minimal resolution and hence it is unique). So, let's blow up a point on the $\pi$-exceptional curve $E\subset Y$: $\sigma: Z\to Y$. Now $Z$ is smooth and $F=\sigma^{_1}_*E$ has self-intersection $F^2=-3$ and $F\simeq \mathbb P^1$, so it can be contracted to a normal surface $Z\to W$. By construction we get a proper birational morphism $W\to X$. This is a partial resolution, but it is not crepant. 

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So, perhaps you mean a partial resolution that's partial to the crepant resolution. I mean that it is an intermediate partial resolution. In the case $n=2$ there is not much room for that, but in general that kind of partial resolutions are also crepant by the following Lemma (but let's do a definition first).

> **Definition** Let $\alpha:X\to Y$ be a proper birational morphism between normal varieties. Assume that $K_Y$ is $\mathbb Q$-Cartier, then   $\alpha$ will be called *crepant* if $K_X\sim \alpha^* K_Y$.

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>**Lemma** Let $\alpha:X\to Y$ and $\beta:Y\to Z$ be two proper birational morphisms and assume that $X,Y,Z$ are normal algebraic surfaces  such that $K_Y$ and $K_Z$ are $\mathbb Q$-Cartier. then if $\beta\circ\alpha$ is crepant, then so are $\alpha$ and $\beta$.



**Proof:** 
Write $$K_X=\alpha^*K_Y+\sum a_iE_i$$ where $E_i$ are $\alpha$-exceptional  irreducible divisors and $$K_Y=\beta^*K_Z+\sum b_jF_j$$ where $F_j$ are $\beta$-exceptional irreducible divisors. It follows that then 
$$K_X=\alpha^*\beta^*K_Z+\sum a_iE_i+\sum b_j\alpha^* F_j.$$
If $\beta\circ\alpha$ is crepant, then $$\sum a_iE_i+\sum b_j\alpha^* F_j\sim 0.$$
However, for each $j$, we have that $\alpha^*F_j= \widetilde F_j +\sum a_{ij}E_i$ where $\widetilde F_j$ is the strict transform of $F_j$ on $X$, so we get that for some other set of coefficients we have
$$\sum c_iE_i+\sum b_j\widetilde F_j\sim 0.$$
The $E_i$ and $\widetilde F_j$ together are all $\beta\circ\alpha$ exceptional and hence linearly independent. Therefore $b_j=0$ for all $j$ and then 
$$\sum a_iE_i\sim 0$$
with the original $a_i$'s. The same reasoning says that then the $a_i=0$ which proves the statement.$\square$

> **Remark**   
It seems to me that the same proof works in arbitrary dimension, so one does not need to limit to surfaces for the Lemma.