Answer: $N = 1$.

Recall that
$$\frac{1}{2^2} + \frac{1}{3^2} + \dots + \frac{1}{n^2} < 1.$$
for all $n > 1$. Using this inequality, show by induction on $N$ that the board may get at most $N^2$ numbers, with equality achieved only for $N = 1$.

Lemma. For any natural $n > 1$, the inequality
$$\frac{1}{2^2} + \frac{1}{3^2} + \dots + \frac{1}{n^2} < 1$$
holds.

Proof. Clearly, for $t > 1$,
$$\frac{1}{t^2} < \frac{1}{t(t-1)} = \frac{1}{t-1} - \frac{1}{t}.$$
Summing such inequalities for $t = 2, 3, \dots, n$, we get
$$\frac{1}{2^2} + \frac{1}{3^2} + \dots + \frac{1}{n^2} < 1 - \frac{1}{2} + \frac{1}{2} + \frac{1}{3} + \dots + \frac{1}{n-1} - \frac{1}{n} = 1 - \frac{1}{n} < 1, \quad \square$$
which is what was required.

Let us prove by induction on $N$ that for any natural $N$, the board can contain at most $N^2$ numbers, and exactly $N^2$ numbers can appear only for $N = 1$. This will yield the answer to the problem.

The base case $N = 1$ is obvious. For the induction step, suppose $N > 1$. Let $1 = d_1 < d_2 < \dots < d_k < d_{k+1} = N$ be all divisors of $N$. After the first replacement, the numbers $d_1, d_2, \dots, d_k$ will be on the board. Mentally divide the board into $k$ parts so that the number $d_j$ is in the $j$-th part. At each subsequent replacement, write the successors of a number in the same part where the number itself was. By the induction hypothesis, in the $j$-th part of the board there can be at most $d_j^2$ numbers. Therefore, the total number of numbers on the board cannot exceed $d_1^2 + d_2^2 + \dots + d_k^2$.

Now note that the numbers $N/d_1 > N/d_2 > \dots > N/d_k > N/d_{k+1}$ are also divisors of $N$. Therefore, they are equal to $d_{k+1}, d_k, \dots, d_2, d_1$ respectively. Taking the lemma into account, we get that the total number of numbers on the board does not exceed:
$$\begin{aligned} d_1^2 + d_2^2 + \dots + d_k^2 &= \frac{N^2}{d_{k+1}^2} + \frac{N^2}{d_k^2} + \dots + \frac{N^2}{d_2^2} = \\ &= N^2 \left( \frac{1}{d_2^2} + \frac{1}{d_3^2} + \dots + \frac{1}{d_{k+1}^2} \right) \le \\ &\le N^2 \left( \frac{1}{2^2} + \frac{1}{3^2} + \dots + \frac{1}{N^2} \right) < N^2 \cdot 1 = N^2, \end{aligned}$$
with strict inequality for $N > 1$.

Therefore, the only possible value is $N = 1$.