China Team Selection Test

Given positive integers $k, m, n$ such that $1 \leq k \leq m \leq n$, we aim to evaluate the sum
$$\sum_{i=0}^n \frac{(-1)^i}{n+k+i} \cdot \frac{(m+n+i)!}{i!(n-i)!(m+i)!}.$$

To solve this, we employ a calculus-based approach. We start by expressing the sum in terms of an integral:
$$\sum_{i=0}^n \frac{(-1)^i}{n+k+i} \cdot \frac{(m+n+i)!}{i!(n-i)!(m+i)!} = \sum_{i=0}^n \frac{(-1)^i (m+n+i)!}{i!(n-i)!(m+i)!} \int_0^1 x^{n+k+i-1} \, dx.$$

This can be rewritten as:
$$\int_0^1 \left[ \sum_{i=0}^n \binom{n}{i} \binom{m+n+i}{n} (-1)^i x^i \right] x^{n+k-1} \, dx.$$

Next, we use the binomial theorem and properties of partial derivatives:
$$\int_0^1 \left[ \sum_{i=0}^n \binom{n}{i} (-1)^i x^i \cdot \frac{1}{n!} \frac{\partial^n}{\partial y^n} y^{m+n+i} \bigg|_{y=1} \right] x^{n+k-1} \, dx.$$

This simplifies to:
$$\frac{1}{n!} \frac{\partial^n}{\partial y^n} \left[ \int_0^1 \sum_{i=0}^n \binom{n}{i} (-1)^i x^{n+k+i-1} y^{m+n+i} \, dx \right]_{y=1}.$$

Further simplification gives:
$$\frac{1}{n!} \frac{\partial^n}{\partial y^n} \left[ \int_0^1 x^{n+k-1} y^{m+n} (1 - xy)^n \, dx \right]_{y=1}.$$

We continue by integrating and differentiating:
$$\frac{1}{n!} \frac{\partial^n}{\partial y^n} \left[ y^{m-k} \int_0^y u^{n+k-1} (1-u)^n \, du \right]_{y=1}.$$

Using the Fundamental Theorem of Calculus, we evaluate:
$$\frac{1}{n!} \sum_{j=0}^n \binom{n}{j} \left[ \frac{\partial^j}{\partial y^j} y^{m-k} \right]_{y=1} \left[ \frac{\partial^{n-j}}{\partial y^{n-j}} \int_0^y u^{n+k-1} (1-u)^n \, du \right]_{y=1}.$$

For $j < n$, the second factor is zero because we differentiate at most $n-1$ times, and there is a zero of order $n$ at $y=1$. For $j=n$, the first factor is zero because $m-k < m \leq n$. Thus, all terms of the sum vanish.

Therefore, the value of the sum is:
$$\sum_{i=0}^n \frac{(-1)^i}{n+k+i} \cdot \frac{(m+n+i)!}{i!(n-i)!(m+i)!} = 0.$$
The answer is: $\boxed{0}.$