1. Given equation

We need to find $n \in \mathbb{N}$ such that

$$\cot^{-1}3+\cot^{-1}4+\cot^{-1}5+\cot^{-1}n=\frac{\pi}{4}.$$

Let

$$A=\cot^{-1}3+\cot^{-1}4+\cot^{-1}5.$$

Then

$$A+\cot^{-1}n=\frac{\pi}{4}.$$

So first we compute $A$.

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2. Use the identity for sum of inverse cotangents

For positive numbers $x,y$,

$$\cot^{-1}x+\cot^{-1}y=\cot^{-1}\!\left(\frac{xy-1}{x+y}\right)$$

whenever the sum lies in $(0,\pi)$, which is true here.

First combine $\cot^{-1}3$ and $\cot^{-1}4$:

$$\cot^{-1}3+\cot^{-1}4 =\cot^{-1}\!\left(\frac{3\cdot4-1}{3+4}\right) =\cot^{-1}\!\left(\frac{11}{7}\right).$$

Now add $\cot^{-1}5$:

$$\cot^{-1}\!\left(\frac{11}{7}\right)+\cot^{-1}5 =\cot^{-1}\!\left(\frac{\left(\frac{11}{7}\right)\cdot5-1}{\frac{11}{7}+5}\right).$$

Simplify:

$$\frac{\frac{55}{7}-1}{\frac{11}{7}+5} =\frac{\frac{48}{7}}{\frac{46}{7}}=\frac{48}{46}=\frac{24}{23}.$$

Thus,

$$A=\cot^{-1}\!\left(\frac{24}{23}\right).$$

So the given equation becomes

$$\cot^{-1}\!\left(\frac{24}{23}\right)+\cot^{-1}n=\frac{\pi}{4}.$$

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3. Convert $\frac{\pi}{4}$ to inverse cotangent form

Since

$$\cot\frac{\pi}{4}=1,$$

we have

$$\frac{\pi}{4}=\cot^{-1}1.$$

Hence,

$$\cot^{-1}\!\left(\frac{24}{23}\right)+\cot^{-1}n=\cot^{-1}1.$$

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4. Take cotangent of both sides using sum formula

Let

$$\alpha=\cot^{-1}\!\left(\frac{24}{23}\right),\qquad \beta=\cot^{-1}n.$$

Then

$$\alpha+\beta=\frac{\pi}{4}.$$

Using

$$\cot(\alpha+\beta)=\frac{\cot\alpha\cot\beta-1}{\cot\alpha+\cot\beta},$$

we get

$$1=\frac{\left(\frac{24}{23}\right)n-1}{\frac{24}{23}+n}.$$

So,

$$\frac{24n}{23}-1 = \frac{24}{23}+n.$$

Multiply by $23$:

$$24n-23=24+23n.$$

Therefore,

$$n=47.$$

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5. Verification

Check:

$$\frac{\left(\frac{24}{23}\right)(47)-1}{\frac{24}{23}+47} =\frac{\frac{1128}{23}-1}{\frac{24}{23}+47} =\frac{\frac{1105}{23}}{\frac{1105}{23}}=1.$$

Hence the sum is indeed $\cot^{-1}1=\frac{\pi}{4}$.

So the required integer is

$$\boxed{47}.$$