Croatia Mathematical Competitions

Since
$$nx = \underbrace{x + \cdots + x}_{n} \quad \text{and} \quad \frac{n(n+3)}{2} = (1 + \cdots + n) + n,$$
the given equation is equivalent to
$$\sum_{k=1}^{n} \frac{(k+1)^2}{x+k} + n x^2 - \sum_{k=1}^{n} x - \sum_{k=1}^{n} k - n = 0,$$
i.e.
$$\sum_{k=1}^{n} \left[ \frac{(k+1)^2}{x+k} - (x+k) \right] + n x^2 - n = 0.$$
After some manipulation we get
$$\frac{(k+1)^2}{x+k} - (x+k) = \frac{(k+1)^2 - (x+k)^2}{x+k} = (1-x) \left(1 + \frac{k+1}{x+k}\right),$$
so the given equation is equivalent to
$$(1-x) \left( n + \sum_{k=1}^{n} \frac{k+1}{x+k} \right) + n(x^2-1) = 0,$$
i.e.
$$(1-x) \left[ n + \left( \frac{2}{x+1} + \frac{3}{x+2} + \dots + \frac{n+1}{x+n} \right) - n(1+x) \right] = 0.$$
For $x = 1$ the equality is obviously satisfied. For $x \ne 1$ we have
$$n + \left( \frac{2}{x+1} + \frac{3}{x+2} + \dots + \frac{n+1}{x+n} \right) = n(1+x),$$
i.e.
$$\frac{2}{x+1} + \frac{3}{x+2} + \dots + \frac{n+1}{x+n} = n x.$$
If $0 < x < 1$, then each of $n$ fractions on the left-hand side is greater than 1, so the left-hand side is greater than $n$, and the right-hand side is less than $n$.
If $x > 1$, exactly the opposite holds: the left-hand side is less than $n$, and the right-hand side is greater than $n$.
Hence, the only solution is $x = 1$.