1. Understanding the Problem:
We need to compute the largest positive integer $k$ such that $5^k$ divides the sum
$$\sum_{d|N} \nu_3(d!)(-1)^{\sigma(d)},$$
where $N = 6^{1999}$.

2. **Analyzing $\sigma(d)$:**
Note that $\sigma(d)$ is odd precisely when $d$ is a power of 2 times an odd square. Since $d$ divides $6^{1999}$, this is equivalent to saying $\nu_3(d)$ is even. Therefore, we have $(-1)^{\sigma(d)} = (-1)^{\nu_3(d) + 1}$.

3. **Rewriting the Sum $S$:**
Let $S$ be the sum in question. We can write:
$$S = \sum_{k=0}^{1999} \sum_{n=0}^{1999} (-1)^{n+1} \nu_3((2^k \cdot 3^n)!).$$

4. Using Legendre's Formula:
Legendre's formula gives:
$$\nu_3((2^k \cdot 3^n)!) = \left\lfloor \frac{2^k \cdot 3^n}{3} \right\rfloor + \left\lfloor \frac{2^k \cdot 3^n}{3^2} \right\rfloor + \left\lfloor \frac{2^k \cdot 3^n}{3^3} \right\rfloor + \cdots.$$
Simplifying, we get:
$$\nu_3((2^k \cdot 3^n)!) = 2^k \left(3^{n-1} + 3^{n-2} + \cdots + 1\right) + \nu_3((2^k)!).$$
This can be further simplified to:
$$\nu_3((2^k \cdot 3^n)!) = 2^{k-1}(3^n - 1) + \nu_3((2^k)!).$$

5. **Summing Over $n$:**
We now sum over $n$:
$$\sum_{n=0}^{1999} (-1)^{n+1} \left(2^{k-1}(3^n - 1) + \nu_3((2^k)!)\right).$$
This simplifies to:
$$2^{k-1} \sum_{n=0}^{1999} (-1)^{n+1} (3^n - 1).$$

6. Using the Geometric Series Formula:
The sum of the geometric series is:
$$\sum_{n=0}^{1999} (-1)^{n+1} 3^n = \frac{3(1 - (-3)^{2000})}{1 - (-3)} = \frac{3^{2000} - 1}{4}.$$
Therefore, the sum becomes:
$$2^{k-1} \cdot \frac{3^{2000} - 1}{4} = 2^k \cdot \frac{3^{2000} - 1}{8}.$$

7. **Summing Over $k$:**
We now sum over $k$:
$$S = \frac{3^{2000} - 1}{8} \sum_{k=0}^{1999} 2^k = \frac{3^{2000} - 1}{8} \cdot (2^{2000} - 1).$$

8. Using the Lifting The Exponent (LTE) Lemma:
Applying the LTE lemma, we get:
$$\nu_5(S) = \nu_5(3^{2000} - 1) + \nu_5(2^{2000} - 1).$$
Using the properties of the LTE lemma:
$$\nu_5(3^{2000} - 1) = \nu_5(81^{500} - 1) = \nu_5(81 - 1) + \nu_5(500) = 1 + 3 = 4,$$
and
$$\nu_5(2^{2000} - 1) = \nu_5(16^{500} - 1) = \nu_5(16 - 1) + \nu_5(500) = 1 + 3 = 4.$$
Therefore,
$$\nu_5(S) = 4 + 4 = 8.$$

The final answer is $\boxed{8}$.