Iberoamerican Olympiad

Given the problem, we want to determine all integers $n \geq 1$ such that the sum of the squares of the digits of $n$, denoted as $s(n)$, is equal to $n$.

To begin, let's express $n$ in terms of its digits. Suppose $n$ is a $k$-digit number given by:
$$n = d_{k-1} \cdot 10^{k-1} + d_{k-2} \cdot 10^{k-2} + \cdots + d_1 \cdot 10^1 + d_0 \cdot 10^0$$
where $d_i$ represents the $i$-th digit of $n$ and $0 \leq d_i \leq 9$ for $0 \leq i \leq k-1$, with $d_{k-1} \neq 0$ because $d_{k-1}$ is the leading digit.

The function $s(n)$ is defined as:
$$s(n) = d_{k-1}^2 + d_{k-2}^2 + \cdots + d_1^2 + d_0^2.$$

Given $s(n) = n$, we want:
$$d_{k-1}^2 + d_{k-2}^2 + \cdots + d_1^2 + d_0^2 = d_{k-1} \cdot 10^{k-1} + d_{k-2} \cdot 10^{k-2} + \cdots + d_1 \cdot 10^1 + d_0.$$

For $k = 1$, $n = d_0$, we reach a straightforward solution:
$$s(n) = d_0^2 = n = d_0 \implies d_0^2 = d_0.$$
Thus, $d_0 = 0$ or $d_0 = 1$. Since $n \geq 1$, we have $d_0 = 1$. This gives $n = 1$.

For $k \geq 2$, we explore the possibility of higher $k$. Notice that for a two-digit number $n$ with digits $d_1, d_0$:
$$s(n) = d_1^2 + d_0^2 \quad \text{and} \quad n = 10d_1 + d_0.$$
Thus, $d_1^2 + d_0^2 = 10d_1 + d_0$.

To analyze further, consider the rough inequality result for the maximum digit value:
$$d_1^2 + d_0^2 \leq 81 + 81 = 162.$$
This suggests $n = 10d_1 + d_0 \leq 162$. However, testing feasible digit options within this limit shows no solutions beyond the trivial single-digit case $n = 1$.

The analysis implies that the only integer $n \geq 1$ satisfying $s(n) = n$ is indeed:
$$\boxed{1}$$
This wraps up the exploration, confirming that 1 is, in fact, the unique solution.