1. Restate the problem and introduce the necessary notation:
Given a positive integer $k$, we call $n$ good if among the binomial coefficients
$$\binom{n}{0}, \binom{n}{1}, \binom{n}{2}, \ldots, \binom{n}{n}$$
at least $0.99n$ of them are divisible by $k$. We need to show that there exists some positive integer $N$ such that among $1, 2, \ldots, N$, there are at least $0.99N$ good numbers.

2. **Generalize the problem for arbitrary constants $0 \le C, D < 1$:**
We will prove that for any constants $0 \le C, D < 1$, there exists an integer $z$ such that for all $N > z$, at least $DN$ of the numbers $1, 2, \ldots, N$ are good, where a number $n$ is good if at least $Cn$ of its binomial coefficients are divisible by $k$.

3. **Prove the statement for $k = p^m$ where $p$ is a prime:**
For a fixed integer $n$ and an integer $k$, the $p$-adic valuation of the binomial coefficient $\binom{n}{k}$ is given by:
$$v_p\left(\binom{n}{k}\right) = \frac{s_p(k) + s_p(n - k) - s_p(n)}{p - 1}$$
where $s_p(x)$ denotes the sum of the digits of $x$ in base $p$. This valuation represents the number of carries when adding $k$ and $n - k$ in base $p$.

4. Lemma:
Let $z \ge 2$ be a positive integer. We can choose $N$ large enough so that when we consider the numbers $1, 2, \ldots, N$ as base-$z$ numbers, at least $DN$ of them have at least $K$ digits among some nonempty subset of digits $S$, where $K$ is any fixed integer.

5. Proof of the Lemma:
For $N$ large enough, we only need to consider the numbers which have at least $x$ digits in base $r$ when written without leading zeroes. We can choose $N$ such that the fraction of numbers with fewer than $x$ digits in base $r$ is as close to $0$ as needed.

Let $q = \frac{|S|}{z}$ and $r = 1 - q$. The probability of a random $x$-digit base-$z$ integer having fewer than $K$ digits among the elements of $S$ is bounded above by:
$$q^x + \binom{x}{1} q^{x-1} r + \cdots + \binom{x}{K-1} q^{x-K+1} r^{K-1} \le \max(q, r)^x \left[ \binom{x}{0} + \binom{x}{1} + \cdots + \binom{x}{K-1} \right]$$
For large $x$, the sum $\binom{x}{0} + \binom{x}{1} + \cdots + \binom{x}{K-1}$ is a polynomial of degree $K-1$ in $x$, denoted by $P(x)$. We need to show that for any $\epsilon > 0$, we can find $x$ such that:
$$\max(q, r)^x P(x) < \epsilon$$
This is equivalent to:
$$\frac{P(x)}{\epsilon} < \left( \frac{1}{\max(q, r)} \right)^x$$
The left-hand side (LHS) is a polynomial in $x$, while the right-hand side (RHS) is exponential in $x$. For large enough $x$, the RHS will dominate the LHS since $\frac{1}{\max(q, r)} > 1$.

6. Application of the Lemma:
As a corollary, we can find $N$ such that among the numbers $1, 2, \ldots, N$, at least $DN$ of them contain at least $K$ zeroes when written in base $p$ without leading zeroes, for some fixed integer $K$.

7. Claim:
For large enough $K$ and $N$, all such numbers are good. Consider a number $n$ which has at least $K$ zeroes and at least $d$ digits in base $p$. The fraction of numbers $0 \le x \le n$ such that $x$ and $n - x$ have at most $A$ carries during their addition can be made arbitrarily small for large $K$ and $d$. If there are no carries at a digit $0$ in $n$, then the corresponding positions in $x$ and $n - x$ must both be $0$. This means that among the $K$ positions, at least $K - A$ of them must be $0$ in $x$. The fraction of such numbers $x$ can be made arbitrarily small when $K$ is large.

8. **Conclusion for $k = p^m$:**
For large enough $d$ and $K$, all these numbers are good. This completes the proof for $k = p^m$.

9. **Generalization to arbitrary $k$:**
Suppose we have proved the statement for $k = a$ and $k = b$, where $a$ and $b$ are coprime. We can find $N$ such that among $1, 2, \ldots, N$, at least $DN$ numbers are good for each $k$. Let $S_1$ and $S_2$ be the sets of good numbers for $k = a$ and $k = b$, respectively, with $|S_i| \ge DN$. The size of their intersection is at least $DN - (1 - D)N = (2D - 1)N$.

For each number $x$ in the intersection, at least $Cx$ of the binomial coefficients are divisible by $a$ and at least $Cx$ are divisible by $b$. The intersection of these sets of binomial coefficients is at least $(2C - 1)x$, and since $a$ and $b$ are coprime, these coefficients are divisible by $ab$. Thus, to prove the statement for $k = ab$, we can choose larger $C$ and $D$ for the statements $k = a$ and $k = b$ to achieve the desired result. This generalizes the proof for all $k \in \mathbb{N}$.

$\blacksquare$