A man has a certain number of small boxes to pack into parcels. If he packs $3$, $4$, $5$ or $6$ in a parcel, he is left with one; if he packs $7$ in a parcel, none is left over. What is the number of boxes he may have to pack?

Let the total number of boxes be $N$.
According to the question, when $N$ is divided by $3$, $4$, $5$, or $6$, it leaves a remainder of $1$.
Thus, the number can be written in the form:

$$N=(\text{LCM of }3,4,5,6)\times k+1$$

First, find the LCM of $3$, $4$, $5$, and $6$:

* $3=3$
* $4=2^2$
* $5=5$
* $6=2\times 3$

$$\text{LCM}=2^2\times 3\times 5=60$$

So, $N=60k+1$.
We are also given that $N$ is completely divisible by $7$ (remainder is $0$ when divided by $7$).
Let us test values of $k$:

* For $k=1$: $N=60(1)+1=61$ (not divisible by $7$)
* For $k=2$: $N=60(2)+1=121$ (not divisible by $7$)
* For $k=3$: $N=60(3)+1=181$ (not divisible by $7$)
* For $k=4$: $N=60(4)+1=241$ (not divisible by $7$)
* For $k=5$: $N=60(5)+1=301$

Let's check if $301$ is divisible by $7$:

$$\frac{301}{7}=43$$

Since $301$ is perfectly divisible by $7$, the number of boxes is $301$.